Undergraduate Lecture Notes in Physics
Reinhard Hentschke
Thermodynamics For Physicists, Chemists and Materials Scient...
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Undergraduate Lecture Notes in Physics
Reinhard Hentschke
Thermodynamics For Physicists, Chemists and Materials Scientists
Undergraduate Lecture Notes in Physics
For further volumes: http://www.springer.com/series/8917
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Series Editors Neil Ashby Professor, Professor Emeritus, University of Colorado Boulder, Boulder, CO, USA William Brantley Professor, Furman University, Greenville, SC, USA Michael Fowler Professor, University of Virginia, Charlottesville, VA, USA Michael Inglis Professor, SUNY Suffolk County Community College, Selden, NY, USA Elena Sassi Professor, University of Naples Federico II, Naples, Italy Helmy S. Sherif Professor, University of Alberta, Edmonton, AB, Canada
Reinhard Hentschke
Thermodynamics For Physicists, Chemists and Materials Scientists
123
Reinhard Hentschke Bergische Universität Wuppertal Germany
ISSN 2192-4791 ISBN 978-3-642-36710-6 DOI 10.1007/978-3-642-36711-3
ISSN 2192-4805 (electronic) ISBN 978-3-642-36711-3 (eBook)
Springer Heidelberg New York Dordrecht London Library of Congress Control Number: 2013938033 Springer-Verlag Berlin Heidelberg 2014 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. Exempted from this legal reservation are brief excerpts in connection with reviews or scholarly analysis or material supplied specifically for the purpose of being entered and executed on a computer system, for exclusive use by the purchaser of the work. Duplication of this publication or parts thereof is permitted only under the provisions of the Copyright Law of the Publisher’s location, in its current version, and permission for use must always be obtained from Springer. Permissions for use may be obtained through RightsLink at the Copyright Clearance Center. Violations are liable to prosecution under the respective Copyright Law. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. While the advice and information in this book are believed to be true and accurate at the date of publication, neither the authors nor the editors nor the publisher can accept any legal responsibility for any errors or omissions that may be made. The publisher makes no warranty, express or implied, with respect to the material contained herein. Printed on acid-free paper Springer is part of Springer Science+Business Media (www.springer.com)
Preface
Many of us associate thermodynamics with blotchy photographs of men in oldfashioned garments posing in front of ponderous steam engines. In fact thermodynamics was developed mainly as a framework for understanding the relation between heat and work and how to convert heat into mechanical work efficiently. Nevertheless, the premises or laws from which thermodynamics is developed are so general that they provide insight far beyond steam engine engineering. Today new sources of useful energy, energy storage, transport, and conversion, requiring development of novel technology, are of increasing importance. This development strongly affects many key industries. Thus, it seems that thermodynamics will have to be given more prominence particularly in the physics curriculum—something that is attempted in this book. Pure thermodynamics is developed, without special reference to the atomic or molecular structure of matter, on the basis of bulk quantities like internal energy, heat, and different types of work, temperature, and entropy. The understanding of the latter two is directly rooted in the laws of thermodynamics—in particular the second law. They relate the above quantities and others derived from them. New quantities are defined in terms of differential relations describing material properties like heat capacity, thermal expansion, compressibility, or different types of conductance. The final result is a consistent set of equations and inequalities. Progress beyond this point requires additional information. This information usually consists in empirical findings like the ideal gas law or its improvements, most notably the van der Waals theory, the laws of Henry, Raoult, and others. Its ultimate power, power in the sense that it explains macroscopic phenomena through microscopic theory, thermodynamics attains as part of Statistical Mechanics or more generally Many-body Theory. The structure of this text is kept simple in order to make the succession of steps as transparent as possible. The first chapter (Two Fundamental Laws of Nature) explains how the first and the second law of thermodynamics can be cast into a useful mathematical form. It also explains different types of work as well as concepts like temperature and entropy. The final result is the differential entropy change expressed through differential changes in internal energy and the various types of work. This is a fundamental relation throughout equilibrium as well as nonequilibrium thermodynamics. The second chapter (Thermodynamic Functions), v
vi
Preface
aside from introducing most of the functions used in thermodynamics, in particular internal energy, enthalpy, Helmholtz, and Gibbs free energy, contains examples allowing to practice the development and application of numerous differential relations between thermodynamic functions. The discussion includes important concepts like the relation of the aforementioned free energies to the second law, extensiveness, and intensiveness as well as homogeneity. In the third chapter (Equilibrium and Stability) the maximum entropy principle is explored systematically. The phase concept is developed together with a framework for the description of stability of phases and phase transitions. The chemical potential is highlighted as a central quantity and its usefulness is demonstrated with a number of applications. The fourth chapter (Simple Phase Diagrams) focuses on the calculation of simple phase diagrams based on the concept of interacting molecules. Here the description is still phenomenological. Equations, rules, and principles developed thus far are combined with van der Waals’ picture of molecular interaction. As a result a qualitative theory for simple gases and liquids emerges. This is extended to gas and liquid mixtures as well as to macromolecular solutions, melts, and mixtures based on ideas due to Flory and others. The subsequent chapter (Microscopic Interactions) explains how the exact theory of microscopic interactions can be combined with thermodynamics. The development is based on Gibbs’ ensemble picture. Different ensembles are introduced and their specific uses are discussed. However, it also becomes clear that exactness usually is not a realistic goal due to the enormous complexity. In the sixth chapter (Thermodynamics and Molecular Simulation) it is shown how necessary and crude approximations sometimes can be avoided with the help of computers. Computer algorithms may even allow tackling problems eluding analytical approaches. This chapter therefore is devoted to an introduction of the Metropolis Monte Carlo method and its application in different ensembles. Thus far the focus has been equilibrium thermodynamics. The last chapter (Non-equilibrium Thermodynamics) introduces concepts in non-equilibrium thermodynamics. The starting point is linear irreversible transport described in terms of small fluctuations close to the equilibrium state. Onsager’s reciprocity relations are obtained and their significance is illustrated in various examples. Entropy production far from equilibrium is discussed based on the balance equation approach and the concept of local equilibrium. The formation of dissipative structures is discussed focusing on chemical reactions. This chapter also includes a brief discussion of evolution in relation to non-equilibrium thermodynamics. There are several appendices. Appendix A: Thermodynamics does not require much math. Most of the necessary machinery is compiled in this short appendix. The reason that thermodynamics is often perceived difficult is not because of its difficult mathematics. It is because of the physical understanding and meticulous care required when mathematical operations are carried out under constraints imposed by process conditions. Appendix B: The appendix contains a listing of a Grand-Canonical Monte Carlo algorithm in Mathematica. The interested reader may use this program to recreate results presented in the text in the context of equilibrium adsorption. Appendix C: This appendix compiles constants, units, and references to useful tables. Appendix D: References are included in the text and as a
Preface
vii
separate list in this appendix. Of course, there are other texts on Thermodynamics or Statistical Thermodynamics, which are nice and valuable sources of information—even if or because some of them have been around for a long time. A selected list is contained in a footnote on page 16. Another listing can be found in the preface to Hill (1986). Wuppertal, Germany
Reinhard Hentschke
Contents
Fundamental Laws of Nature . . . . . . Types of Work . . . . . . . . . . . . . . . . . The Postulates of Kelvin and Clausius . Carnot’s Engine and Temperature . . . . Entropy . . . . . . . . . . . . . . . . . . . . . .
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1 1 15 16 22
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27 27 29 54 68
3
Equilibrium and Stability. . . . . . . . . . . . . . . . . . . . 3.1 Equilibrium and Stability via Maximum Entropy. 3.2 Chemical Potential and Chemical Equilibrium . . 3.3 Applications Involving Chemical Equilibrium . . .
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73 73 80 90
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Simple Phase Diagrams . . . . . . . . . . . . . . . . . . 4.1 Van Der Waals Theory . . . . . . . . . . . . . . . 4.2 Beyond Van Der Waals Theory. . . . . . . . . . 4.3 Low Molecular Weight Mixtures. . . . . . . . . 4.4 Phase Equilibria in Macromolecular Systems
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125 125 140 155 164
5
Microscopic Interactions . . . . . . . . . . . 5.1 The Canonical Ensemble . . . . . . . . 5.2 Generalized Ensembles. . . . . . . . . . 5.3 Grand-Canonical Ensemble. . . . . . . 5.4 The Third Law of Thermodynamics.
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173 173 201 205 217
6
Thermodynamics and Molecular Simulation . 6.1 Metropolis Sampling . . . . . . . . . . . . . . . 6.2 Sampling Different Ensembles . . . . . . . . 6.3 Selected Applications . . . . . . . . . . . . . . .
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221 221 225 227
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Two 1.1 1.2 1.3 1.4
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Thermodynamic Functions . . . . . . . . . 2.1 Internal Energy and Enthalpy . . . . 2.2 Simple Applications. . . . . . . . . . . 2.3 Free Energy and Free Enthalpy . . . 2.4 Extensive and Intensive Quantities
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ix
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Contents
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239 240 250 263 273
Appendix A: The Mathematics of Thermodynamics . . . . . . . . . . . . . .
281
Appendix B: Grand-Canonical Monte Carlo: Methane on Graphite . . . . . . . . . . . . . . . . . . . . . . . . . .
289
Appendix C: Constants, Units, Tables . . . . . . . . . . . . . . . . . . . . . . . .
293
Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
299
7
Non-Equilibrium Thermodynamics . . . 7.1 Linear Irreversible Transport. . . . . 7.2 Entropy Production . . . . . . . . . . . 7.3 Complexity in Chemical Reactions 7.4 Remarks on Evolution . . . . . . . . .
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Chapter 1
Two Fundamental Laws of Nature
1.1 Types of Work 1.1.1 Mechanical Work A gas confined to a cylinder absorbs a certain amount of heat, δq. The process is depicted in Fig. 1.1. According to experimental experience this leads to an expansion of the gas. The expanding gas moves a piston to increase its volume by an amount δV = Vb − Va . For simplicity we assume that the motion of the piston is frictionless and that its mass is negligible compared to the mass, m, of the weight pushing down on the piston. We do not yet have a clear understanding of what heat is, but we consider it a form of energy which to some extend can be converted into mechanical work, w.1 In our case this is the work needed to lift the mass, m, by a height, δs, against the gravitational force m g. From mechanics we know δwdone by gas =
b
ds · fgas = −
a
= Pex
b
ds · m g
a Vb
d V = Pex δV.
Va
Here Pex = mg/A is the external pressure exerted on the gas due to the force mg acting on the cross-sectional area, A (δV = Aδs). The process just described leads to a change in the total energy content of the gas, δ E. The gas receives a positive amount of heat, δq. However, during the expansion it also does work and thereby reduces its total energy content, in the following called 1
Originally it was thought that heat is a sort of fluid and heat transfer is transfer of this fluid. In addition, it was assumed that the overall amount of this fluid is conserved. Today we understand that heat is a form of dynamical energy due to the disordered motion of microscopic particles and that heat can be changed into other forms of energy. This is what we need to know at this point. The microscopic level will be addressed in Chap. 5. R. Hentschke, Thermodynamics, Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-36711-3_1, © Springer-Verlag Berlin Heidelberg 2014
1
2
1 Two Fundamental Laws of Nature
mg
b mg
a
s
q
Fig. 1.1 A gas confined to a cylinder absorbs a certain amount of heat, δq
internal energy, by −Pex δV . The combined result is δ E = δq − Pex δV. Notice that after the expansion has come to an end we have Pex = P, where P is the gas pressure inside the cylinder. In particular we know that P is a function of the volume, V , occupied by the gas, i.e. P = P(V ). In the following we assume that the change in gas pressure during a small volume change δV is a second order effect which can be neglected. Therefore for small volume changes we have δ E = δq − PδV.
(1.1)
This is the first law of thermodynamics for this special process. It uses energy conservation to distinguish the different contributions to the total change in internal energy of a system (here the gas) during a thermodynamic process (here absorption of heat plus volume expansion). We just have introduced two important concepts frequently used in thermodynamics—process and system. The latter requires the ability to define a boundary between “inside” and “outside”. Both, the inside and the outside, may be considered systems individually. Systems usually are distinguished according to their degree of openness. Isolated system means that this system exchanges nothing with its exterior. An open system on the other hand may exchange everything there is to exchange, like heat or matter. A closed system holds back matter but allows heat exchange, e.g. the above gas filled cylinder. Systems are sometimes divided into subsystems. Subsystems, however, are still systems. After having defined or (better) prepared a system we may observe what happens to it or we may actively do something to it. This “what
1.1 Types of Work
3
happens to it” or “doing something” means that the system undergoes a process (of change). A special type of system is the reservoir. A reservoir usually is in thermal contact with our system of interest. Thermal contact means that heat may be transferred between the reservoir and our system of interest. However, the reservoir is so large that there is no measurable change in any of its physical properties due to the exchange. Now we proceed replacing the above gas by an elastic medium. Those readers who are not sufficiently familiar with the theory of elastic bodies may skip ahead to “Electric work” (p. 7).
Mechanical Work Involving Elastic Media We consider an elastic body composed of volume elements d V depicted in Fig. 1.2. The total force acting on the elastic body may be calculated according to V
d V fα
(1.2)
for every component α (= 1, 2, 3 or x, y, z). Here f is a force density, i.e. force per volume. Assuming that the f α are purely elastic forces acting between the boundaries of the aforementioned volume elements inside V , i.e. excluding for instance gravitational forces or other external fields acting on volume elements inside the elastic body, we may define the internal stress tensor, σ , via fα =
3 ∂σαβ ∂σαβ ≡ . ∂ xβ ∂ xβ
(1.3)
β=1
Fig. 1.2 Elastic body composed of volume elements d V
dV
V
4
1 Two Fundamental Laws of Nature
x
shear force on -face / area V
normal force on -face / area x Fig. 1.3 The relation between indices, force components, and the faces of the cubic volume element
Here we apply the summation convention, i.e. if the same index appears twice on the same side of an equation then summation over this index is implicitly assumed (unless explicitly stated otherwise). The relation between indices, force components, and the faces of the cubic volume element is depicted in Fig. 1.3. Upper and lower sketches illustrate the shear and the normal contribution to the force component f α acting on the volume element in α-direction. Notice that f α can be written as the sum over two shear stress and one normal stress contribution. The latter are stress differences between adjacent faces of the cubic volume element. Note also that the unit of σαβ is force per area. We want to calculate the work δw done by the f α during attendant small displacements δu α , i.e. ∂σαβ (1.3) δw = d V f α δu α = dV δu α . ∂ xβ The integral may be rewritten using Green’s theorem in space: δw =
σαβ δu α d Aβ −
d V σαβ
∂δu α . ∂ xβ
1.1 Types of Work
5
We neglect the surface contribution2 and use the symmetry property of the stress tensor3 to obtain ∂u ∂u β
∂δu α 1 α . δw = − d V σαβ =− + d V σαβ δ ∂ xβ 2 ∂ xβ ∂ xα ∼ =2u αβ
The quantity u αβ is the strain tensor (here for small displacements). The final result is δw = −
d V σαβ δu αβ .
(1.4)
We want to work this out in three simple cases. First we consider a homogeneous dilatation of a cubic volume V = L x L y L z . We also assume that the shear components of the stress tensor vanish, i.e. σαβ = 0 for α = β. In such a system the normal components of the stress tensor should all be the same, i.e. σ ≡ σx x = σ yy = σzz . We thus have σαβ δu αβ = σx x δu x x + σ yy δu yy + σzz δu zz = σ (δu x x + δu yy + δu zz ).
(1.5)
Homogeneous deformation means δL α ∂u α = . ∂ xα Lα
(1.6)
And because u αα = ∂u α /∂ xα (no summation convention here) we obtain σαβ δu αβ = σ
δL y δL x δL z + + Lx Ly Lz
=σ
δV . V
(1.7)
Integration over the full volume then yields
2
For a discussion see Landau et al. (1986). To show the symmetry of the stress tensor, i.e. σαβ = σβα , we compute the torque exerted by the f α in a particular volume element integrated over the entire body: ∂σβγ ∂σαγ d V ( f α xβ − f β xα ) = d V xβ − xα ∂ xγ ∂ xγ ∂(σαγ xβ − σβγ xα ) = dV − d V (σαγ δβγ − σβγ δαγ ) ∂ xγ = (σαγ xβ − σβγ xα )d Aγ − d V (σαβ − σβα ).
3
The volume integral must vanish in order for the net torque to be entirely due to forces applied to the surface of the body.
6
1 Two Fundamental Laws of Nature
δw = −σ δV,
(1.8)
i.e. we recover the above gas case with P = −σ . In a second example we consider the homogeneous dilatation of a thin elastic sheet. The sheet’s volume is V = Ah = L x L y h, where the thickness, h, is small and constant. Now we have δL y δA δL x =σ (1.9) + σαβ δu αβ = σ Lx Ly A and therefore δw = −σ hδ A ≡ −γ δ A.
(1.10)
The quantity γ is the surface tension. An obvious third example is the homogeneous dilatation of a thin elastic column V = h 2 L z . Here h 2 is the column cross sectional area and L z is its length. This time we have δL z (1.11) σαβ δu αβ = σ Lz and thus δw = −σ AδL z ≡ −T δL z ,
(1.12)
where T is the tension. Example: Expanding Gas. We consider the special case of the first law expressed in Eq. (1.1). If we include the surface tension contribution to the internal energy of the expanding gas, then the resulting equation is δ E = δq − PδV + γ δ A.
(1.13)
We remark that the usual context in which one talks about surface tension refers to interfaces. This may be the interface between two liquids or the surface of a liquid film relative to air, e.g. a soap bubble. In the latter case there are actually two surfaces. In such cases we define γ = f T /(2l), which reflects the presence of two surfaces. Example: Fusing Bubbles. An application of surface tension is depicted in Fig. 1.4. The figure depicts two soap bubbles touching and fusing. We ask whether the small bubble empties its gas content into the large one or vice versa. We may answer this question by considering the work done by one isolated bubble during a small volume change:
1.1 Types of Work
7
Fig. 1.4 An application of surface tension
δwdone by gas in bubble = Pex δV + γ δ A. Notice that the sign of the surface tension contribution has changed compared to Eq. (1.10). This is because in Eq. (1.10) we compute the work done by the membrane. But here the gas is doing work on the membrane, which changes the sign of this work contribution. The same work, i.e. δwdone by gas in bubble , can be written in terms of the pressure, P, inside the bubble, δwdone by gas in bubble = PδV. Combining the two equations and using δV = 4πr 2 δr and δ A = 8πr δr , where r is the bubble radius, yields P = Pex +
2γ . r
We conclude that the gas inside the smaller bubble has the higher pressure and therefore the smaller bubble empties itself into the larger bubble.
1.1.2 Electric Work We now consider work involving electric and magnetic variables4,5 starting with an example.
4 5
Here we use Gaussian units. The conversion to SI-units is tabulated in Appendix C. Three early but very basic papers in this context are: Guggenheim (1936a, b); Koenig (1937).
8
1 Two Fundamental Laws of Nature
Example: Charge Transfer Across a Potential Drop. A charge δq in an electric field experiences the force F = δq E (Do not confuse this δq with the previously introduced heat change!). Consequently the work done by the charge-field system if the charge moves from point a to point b in space is
b
δwq = a
ds · F = −δq
b
ds · ∇φ.
(1.14)
a
and thus Here φ is the potential, i.e. E = −∇φ, δwq = −δqφba ,
(1.15)
where φba = φ(b) − φ(a) is the potential difference between b and a. The corresponding internal energy of the charge-field system changes by δ E q = −δwq = δqφba .
(1.16)
This equation may be restated for a charge current I = δq/δt, where δt is a certain time interval: (1.17) δ E I = I φba δt. In the presence of the resistance, R, the quantity δq J oule = R I 2 δt is the Joule heat generated by the current (James Prescott Joule, British physicist, *Salford (near Manchester) 24.12.1818, †Sale (County Cheshire) 11.10.1889; made important contributions to our understanding of heat in relation to mechanical work (Joule heat) and internal energy (Joule-Thomson effect).). Now we consider the following equations appropriate for continuous dielectric media: + 4π j × H = 1 ∂ D ∇ (1.18) c ∂t c and
× E = − 1 ∂ B. ∇ c ∂t
(1.19)
The second equation simply follows by the usual spatial averaging procedure applied r ) is the average electric to the corresponding vacuum Maxwell’s equation.6 Here E( field in a volume element at point r. This volume element is large compared to atomic is the displacement field given by D = E + 4π P. dimensions. In the same sense D P is the macroscopic polarization, i.e. the local electrical dipole moment per volume. is the average magnetic field (magnetic induction), Analogously B = H + 4π M 6
James Clerk Maxwell, British physicist, *Edinburgh 13.6.1831, †Cambridge 5.11.1879; particularly known for his unified theory of electromagnetism (Maxwell equations).
1.1 Types of Work
9
is the macroscopic magnetization, i.e. the local magnetic dipole moment per and M volume. The first equation is less obvious and requires a more detailed discussion. We consider a current density je inside a medium due to an extra (“injected”) charge density ρe . The two quantities fulfill the continuity equation ∂ρe · je = 0. +∇ ∂t We also have
= 4πρe . ·D ∇
Differentiation of this Maxwell equation with respect to time and inserting the result into the previous equation yields
· ∇
∂D + 4π je ∂t
= 0.
The expression in brackets is a vector, which may be expressed as the curl of another vector c H , i.e. × H = 1 ∂ D + 4π je . ∇ c ∂t c Comparison of this with Ampere’s law in vacuum suggests indeed H = H and c = c. We thus arrive at Eq. (1.18). An in depths discussion can be found in Lifshitz et al. (2004). We proceed by multiplying Eq. (1.18) with c E/(4π ) and Eq. (1.19) with −c H /(4π ). Adding the two equations yields
1 ∂ B c c 1 ∂D + + j · E. E · ∇ × H − H · ∇ × E = E· H· 4π 4π 4π ∂t 4π ∂t = b · (∇ this is · ( × a ) − a · (∇ × b) With the help of the vector identity ∇ a × b) transformed into
c 1 ∂ B 1 ∂D + + j · E. ∇ · H × E = E· H· 4π 4π ∂t 4π ∂t Now we integrate both sides over the volume V and use Green’s theorem in space = · ( H × E) (also called divergence theorem), i.e. V d V ∇ A d A · ( H × E), where A is a surface element on the surface A of the volume oriented towards the outside of V . If we choose the volume so that the fields vanish on its surface, then · ( H × E) = 0 (The configuration of the system is fixed during all of this.). d A A Thus our final result is
10
1 Two Fundamental Laws of Nature
Fig. 1.5 A cylindrical volume element whose axis is parallel to j
A j
dV
δ B δD + H · + j · Eδt E · 4π 4π
s
= 0.
(1.20)
The third term in Eq. (1.20) is the work done by the E-field during the time δt. To see this we imagine a cylindrical volume element whose axis is parallel to j depicted in Fig. 1.5. Then δV = Aδs and δV j = (q/δt)δs , where q is the charge passing = q E · δs , where q E is the through the area A during the time δt. Thus δV j · Eδt force acting on the charge q doing work (cf. the above example). We conclude that we may express the work done by the system, δw = d V j· E δt, by the other two terms in Eq. (1.20) describing the attendant change of the electromagnetic energy content of the system. For a process during which the system exchanges heat and is doing electrical work we now have
δ E = δq +
dV
δ B δD + H · E · 4π 4π
.
(1.21)
D, B, and H are more difficult to deal with than fields in vacuum. The quantities E, Nevertheless, for the moment we postpone a more detailed discussion and return to Eq. (1.21) on p. 57.
1.1.3 Chemical Work As a final example consider an open system—one we can add material to. Generally, work must be done to increase the amount of material in a system. The work done depends on the state of the system. If we add δn moles of material,7 we write the
One mole (n = 1) is an amount of substance of a system which contains as many elementary units as there are atoms of carbon in 12 g of the pure nuclide carbon-12. The elementary unit may be an atom, molecule, ion, electron, photon, or a specified group of such units.
7
1.1 Types of Work
11
work done on the system as δwdone on system = μδn.
(1.22)
The quantity μ is called the chemical potential (per mole added). In a more general situation a system may contain different species. We shall say that these are different components i. Now the above equation becomes δwdone on system =
μi δn i .
(1.23)
i
Here μi is the chemical potential of component i. Thus for a process involving exchange of heat as well as chemical work we have δ E = δq +
μi δn i .
(1.24)
i
1.1.4 The First Law The first law is expressing conservation of energy. The specific terms appearing in the first law do depend on the types of work occurring in the process of interest. The following box contains a number of examples. Example: Statements of the First Law for Different Processes. (i)
δ E = δq − PδV + γ δ A +
μi δn i
i
This describes a process during which heat is exchanged by the system and its exterior. Mechanical work in the form of volume work and surface work is done in addition. The composition of the system changes as well. (ii)
δ E = δq −
d V j · Eδt
Here the process of interest involves heat exchange and electrical work.
12
1 Two Fundamental Laws of Nature
(iii)
δ E = δq − PδV +
dV
+ H · δ B E · δ D 4π
This example is for a process during which heat is exchanged and both volume and electrical work is done. More generally the first law is expressed via δ E = δq − δw.
(1.25)
However, there is an alternative sign convention used in some of the literature, i.e. δ E = δq + δw.
(1.26)
The sign preceding δw depends on the meaning of the latter. In Eq. (1.25) δw always is the work done by the system for which we write down the change in the system’s internal energy, δ E, during a process involving both heat transfer and work. In Eq. (1.26) on the other hand δw is understood as work done on the system. In the following we shall use the sign convention as expressed in Eq. (1.25)! Another point worth mentioning is the usage of the symbols δ, , and d. δ denotes a small change (afterwards–before) during a process. basically has the same meaning, except that the change is not necessarily small. Even though d indicates a small change just like δ, it has an additional meaning—indicating exact differentials. This is something we shall discuss in much detail latter in the text. But for the benefit of those who compare the form of Eq. (1.25) to different texts, we must add a provisional explanation. In principle every process has a beginning and an end. Beginning and end, as we shall learn, are defined in terms of specific values of certain variables (e.g. values of P and V ). These two sets of variable values can be connected by different processes or paths in the space in which the variables “live”. If a quantity changes during a process and this change only depends on the two endpoints of the path rather than on the path as a whole, then the quantity possesses an exact differential and vice versa. In the case of mechanical work, for instance, we can imagine pushing a cart from point A to point B. There may be two alternative routes—one involving a lot of friction and a “smooth” one causing less friction. In the former case one may find Eq. (1.25) stated as d E = δq − δw. (1.27) This form explicitly distinguishes between the exact differential d E and the quantities δq and δw, which are not exact differentials. In the case of δw this is in accord with our cart-pushing example, because the work does depend on the path we choose. For the two other quantities we shall show their respective property latter in this text (cf. p. 283ff), when we deal with the mathematics of exact differentials.
1.1 Types of Work
13
However, already at this point we remark that the expressions we have derived in our examples for the various types of work will reappear with d instead of δ. This is because we focus on what we shall call reversible work. Friction, occurring in the cart-pushing example or possibly in Fig. 1.1 when the gas moves the piston, is neglected as well as other types of loss. The following is an example illustrating what we mean by reversible vs. irreversible work. Example: Reversible and Irreversible Work. In an isotropic elastic body the following equation holds (Landau et al. 1986):
α=β
σαβ = 2μu αβ = μ
∂u β ∂u α + ∂ xβ ∂ xα
.
(1.28)
On the right is the stress tensor and on the left the product of 2μ with the strain tensor (for small strain). The quantity μ is the shear modulus (not to be confused with the chemical potential). This equation is related to the two upper sketches in Fig. 1.3. If in the depicted situation (shear force acting on β-face is applied in α-direction) there is little or ideally no strain in β-direction (this is like shearing a deck of cards), then the above equation may be written as σμ ≡ σαβ = μ
∂u α ≡ μu μ . ∂ xβ
(1.29)
Real shear is accompanied by friction. Experience suggests that friction often can be described by an equation akin to the above: ση = ηu˙ η .
(1.30)
The quantity η is a friction coefficient and u˙ η is a strain rate. Figure 1.6, showing a spring and a dashpot, is a pictorial representation of Eqs. (1.29) and (1.30). Figure 1.7 shows three simple combinations (a, b, c) of the two elements depicted in Fig. 1.6. These combinations may be translated into differential equations and serve as simple models for so called viscoelastic behavior (Wrana 2009). Important viscoelastic materials are the tread compounds in automobile tires. In the following we merely focus on sketch (a). Its translation is (u ≡ u μ = u η ). (1.31) σ ≡ σμ + ση = μu + ηu˙ We assume that the applied stress is σ = σo sin(ωt +δ), where ω is a frequency, t is time, and δ is a phase. The attendant strain is u = u o sin(ωt) (This is a simple mathematical description of an experimental procedure in what is called
14
1 Two Fundamental Laws of Nature
Fig. 1.6 Pictorial representation of Eqs. (1.29) and (1.30)
µ
Fig. 1.7 Three simple combinations (a, b, c) of the two elements
µ
(a)
(b)
µ (c)
µ
1
µ
2
dynamic mechanical analysis.). Inserting this into Eq. (1.31) we find the relations σo cos δ = μu o ≡ μ u o σo sin δ = ηωu o ≡ μ u o .
(1.32) (1.33)
The two newly defined quantities μ and μ are called storage and loss modulus, respectively. Their meaning becomes clear if we compute the work done during one full shear cycle, i.e.
σ du = 0
2π/ω
σ udt ˙ = π μ u 2o .
(1.34)
1.1 Types of Work
15
Actually this is work per volume, cf. (1.4). However, if we do the same calculation just for the first quarter cycle (form zero to maximum shear strain) the result is π/(2ω) 1 1 σ udt ˙ = μ u 2o + π μ u 2o . (1.35) 2 4 0 The first term is the reversible part of the work, which does not contribute to the integral in the case of a full cycle. This term is analogous to the elastic energy stored in a stretched/compressed spring. The second term as well as the result in Eq. (1.34) cannot be recovered and is lost, i.e. producing heat. Models like ours only convey a crude understanding of loss or dissipative processes in viscoelastic materials. Considerable effort is spend by the R&D departments of major tire makers to understand and control loss on a molecular basis. In tire materials the moduli themselves strongly depend on the shear amplitude. Understanding and controlling this effect, the Payne effect, is one important ingredient for the improvement of tire materials, e.g. optimizing rolling resistance (Vilgis et al. 2009).
1.2 The Postulates of Kelvin and Clausius The first law does not address the limitations of heat conversion into work or heat transfer between systems. The following two postulates based on experimental experience do just this. They are the foundation of what is called the second law of thermodynamics.8
1.2.1 Postulate of Lord Kelvin (K) A complete transformation of heat (extracted from a uniform source) into work is impossible.9
8
Here we follow Fermi (1956). Dover (Enrico Fermi, Nobel prize in physics for his contributions to nuclear physics, 1938). 9 Thomson, Sir (since 1866) William, Lord Kelvin of Largs, (since 1892), British physicist, *Belfast 26.6.1824, †Netherhall (near Largs, North Ayrshire) 17.12.1907; one of the founders of classical thermodynamics; among his achievements are the Kelvin temperature scale, the discovery of the Joule-Thomson effect in 1853 with J. P. Joule and the thermoelectric Thomson effect in 1856, as well as the development of an atomic model with J. J. Thomson in 1898.
16
1 Two Fundamental Laws of Nature
1.2.2 Postulate!of Clausius (C) It is impossible to transfer heat from a body at a given temperature to a body at higher temperature as the only result of a transformation.10 Remark: At this point we use the “temperature” θ to characterize a reservoir as hotter or colder than another. The precise meaning of temperature is discussed in the following section. These two postulates are equivalent. A way to prove this is by assuming that the first postulate is wrong. This is then shown to contradict the second postulate. Subsequently the same reasoning is applied starting with the second postulate, i.e. the assumption that the second postulate is wrong is shown to contradict the first. First we assume (K) to be false. Figure 1.8 illustrates what happens. At the top is a reservoir at a temperature θ1 surrendering heat q to a device (circle) which converts this exact amount of heat into work w. A process possible if (K) is false. At the bottom this setup is extended by a friction device (f) converting the work w into heat q, which is transferred to a second reservoir at θ2 (> θ1 ). Thus the only overall result of the process is the transfer of heat from the colder to the hotter reservoir. We therefore contradict (C). Now we assume that (C) is false. The upper part of Fig. 1.9 shows heat q flowing from the colder reservoir to the hotter reservoir—with no other effect. At the bottom this setup is extended. The heat q is used to do work leaving the upper reservoir unaltered. Clearly, this is in violation of (K). Therefore both postulates are equivalent. They have important consequences, which we explore below.
1.3 Carnot’s Engine and Temperature Consider a fluid undergoing a cyclic transformation shown in Fig. 1.10. The upper graph shows the cycle in the P-V -plane, whereas the lower is a sketch illustrating the working principle of a corresponding device. Here the amount of heat q2 is transferred from a heat reservoir at temperature θ2 (θ2 > θ1 ) to the device. During the transfer (path from a to b in the P-V -diagram) the temperature in the device is θ2 . This part of the process is an isothermal expansion. Then the device crosses via adiabatic11 expansion to a second isotherm at temperature θ1 , the temperature of a 10
Rudolf Julius Emanuel Clausius, German physicist, *Köslin (now Koszalin) 2.1.1822, †Bonn 24.8.1888; one of the developers of the mechanical theory of heat; his achievements encompass the formulation of the second law and the introduction of the “entropy” concept. 11 A transformation of a thermodynamic system is adiabatic if it is reversible and if the system is thermally insulated. Definitions of an adiabatic process taken from the literature: Pathria (1972): “Hence, for the constancy of S (Entropy) and N (number of particles), which defines an adiabatic process, ....”
1.3 Carnot’s Engine and Temperature
17
Fig. 1.8 Assumption of postulates: Kelvin
w
q 1 2
q w
f
q 1
Fig. 1.9 Assumption of postulates: Clausius
2
q
q 1
2
q
q w
q
q' 1
18
1 Two Fundamental Laws of Nature
P a q=0
q2 b
1
2
d
q=0 c
q1
V 2
q2 w
C q1 1 Fig. 1.10 Fluid undergoing a cyclic transformation
Fermi (1956): “A transformation of a thermodynamic system is said to be adiabatic if it is reversible and if the system is thermally insulated so that no heat can be exchanged between it and its environment during the transformation” Pauli (1973): “Adiabatic: During the change of state, no addition or removal of heat takes place; ....” Chandler (1987): “ ...the change S is zero for a reversible adiabatic process, and otherwise S is positive for any natural irreversible adiabatic process.” Guggenheim (1986): “When a system is surrounded by an insulating boundary the system is said to be thermally insulated and any process taking place in the system is called adiabatic. The name adiabatic appears to be due to Rankine (Maxwell, Theory of Heat, Longmans 1871).” Kondepudi and Prigogine (1998): “In an adiabatic process the entropy remains constant.” We note that for some authors “adiabatic” includes “reversibility” and for others, here Pauli, Chandler, and Guggenheim, “reversibility” is a separate requirement, i.e. during an “adiabatic” process no heat change takes place but the process is not necessarily reversible. (see also the discussion of the “adiabatic principle” in Hill (1956).)
1.3 Carnot’s Engine and Temperature
19
Fig. 1.11 Proof of Carnot’s theorem
2
q2 C q1
q'2 w
w'
X q'1 1
second reservoir (path from b to c in the P-V -diagram).12 Now follows an isothermal compression during which the device releases the amount of heat q1 into the second reservoir (path from c to d in the P-V -diagram). The final part of the cycle consists of the crossing back via adiabatic compression to the first isotherm (path from d to a in the P-V -diagram). In addition to the heat transfer between reservoirs the device has done the work w. Any device able to perform such a cyclic transformation in both directions is called a Carnot engine.13,14 According to the first law, δ E = δq − δw, applied to the Carnot engine we have E = 0 and thus w = q2 −q1 . Our Carnot engine has a thermal efficiency, generally defined by work done w (1.36) η= = , heat absorbed q2 which is η =1−
q1 . q2
(1.37)
Remark: If the arrows in Fig. 1.10 are reversed the result is a heat pump, i.e. a device which uses work to transfer heat from a colder reservoir to a hotter reservoir. The efficiency of such a device is 1/η. Here the aim is to use as little work as possible to transfer as much heat as possible. Now we prove an interesting fact—the Carnot engine is the most efficient device, operating between two temperatures, which can be constructed! This is called Carnot’s theorem. To prove Carnot’s theorem we put the Carnot engine (C) in series with an arbitrary competing device (X) as shown in Fig. 1.11.
12
Do you understand why the slopes of the isotherms are less negative than the slopes of the adiabatic curves? You find the answer on p. 40. 13 Nicolas Léonard Sadi Carnot, French physicist, *Paris 1.6.1796, †ibidem 24.8.1832; his calculations of the thermal efficiency for steam engines prepared the grounds for the second law. 14 If you are interested in actual realizations of the Carnot engine and what they are used for visit http://www.stirlingengine.com.
20
1 Two Fundamental Laws of Nature
First we note that if we operate both devices many cycles we can make their total heat inputs added up over all cycles, q2 and q2 , equal (i.e., q2 = q2 with arbitrary precision). After we have realized this we now reverse the Carnot engine (all arrows on C are reversed). Again we operate the two engines for as many cycles as it takes to fulfill q2 = q2 . This means that reservoir 2 is completely unaltered. But what are the consequences of all this? According to the first law we have 1.law
wtotal = q2,total − q1,total
(1.38)
where q2,total = −q2 + q2 = 0 q1,total = −q1 + q1 . Because the second reservoir is unaltered we must have wtotal ≤ 0.
(1.39)
wtotal > 0 violates Kelvin’s postulate! However, this implies q1,total ≥ 0 q1 ≥ q1 q1 q2 ≥ q1 q2 And therefore ηX = 1 −
q1 q1 ≥ . q2 q2
q1 q1 ≤1− = ηCar not . q2 q2
(1.40)
There is no device more efficient than Carnot’s engine. Question: Do you understand what distinguishes the Carnot engine in this proof from its competitor? It is the reversibility. If the competing device also is fully reversible we can redo the proof with the two engines interchanged. We then find ηCar not ≤ η X , and thus ηCar not = η X . We may immediately conclude the following corollary: All Carnot engines operating between two given temperatures have the same efficiency. This in turn allows to define a temperature scale using Carnot engines. The idea is illustrated in Fig. 1.12. We imagine a sequence of Carnot engines all producing the same amount of work w. Each machine uses the heat given off by the previous engine as input. According to the first law w = qi+1 − qi .
(1.41)
1.3 Carnot’s Engine and Temperature
21
Fig. 1.12 Defining temperature scale using Carnot engines
qi+1 C
w
qi qi C
w
qi-1
We define the reservoir temperature θi via θi = xqi ,
(1.42)
where x is a proportionality constant independent of i. Thus the previous equation becomes (1.43) xw = θi+1 − θi . We may for instance choose xw = 1K , i.e. the temperature difference between reservoirs is 1K . We remark that this definition of a temperature scale is independent of the substance used. Furthermore the thermal efficiency of the Carnot engine becomes ηCar not = 1 −
θ1 θ2
(1.44)
(θ2 > θ1 ). Notice that the efficiency can be increased by making θ1 as low and θ2 as high as possible. Notice also that θ1 = 0 is not possible, because this violates the second law. θ1 can be arbitrarily close but not equal to zero. On p. 42 we compute the thermal efficiency for the Carnot cycle in Fig. 1.10 using an ideal gas as working medium. We shall see that for the ideal gas temperature T ∝ θ . Thus from here on we use θ = T .
22
1 Two Fundamental Laws of Nature
System q2 T1
T2
T3
Tn
q1
q2
q3
qn
C
w1
q1,0
C
C
w2
q2,0
w3
q3,0
C
wn
qn,0 T0
Fig. 1.13 Use of the assembly of Carnot engines and reservoirs
1.4 Entropy Some of you may have heard about the thermodynamic time arrow. Gases escape from open containers and heat flows from a hot body to its colder environment. Never has spontaneous reversal of such processes been observed. We call these irreversible processes. The world is always heading forward in time. Mathematically this is expressed by Clausius’ theorem.
1.4.1 Theorem of Clausius In any cyclic transformation throughout which the temperature is defined, the following inequality holds dq ≤ 0. (1.45) T The integral extends over one cycle of the transformation. The equality holds if the cyclic transformation is reversible. Proof: We make use of the assembly of Carnot engines and reservoirs shown in Fig. 1.13. The device called system successively visits all reservoirs indicated by the temperatures T1 to Tn . After it has visited reservoir Tn it is in the same state as in the beginning.15 According to Eq. (1.42) we may write 15
To achieve this not all Carnot engines operate in the same direction.
1.4 Entropy
23
qi,0 =
T0 qi . Ti
Thus the total heat surrendered by the reservoir at T0 (T0 > Ti and i = 1, 2, ..., n) in one complete turn around of the system is q0 =
n
qi,0 = T0
i=1
n qi . Ti i=1
As before, when we compared the thermal efficiency of the Carnot engine to the X-machine (on p. 19), we use the first law, i.e. 0 = E =
n
qi,0 −
i=1
=q0
n
qi −wtotal .
i=1
=0
Because wtotal ≤ 0 if Kelvin’s postulate is correct, we must have q0 ≤ 0 and consequently n qi ≤ 0. Ti i=1
Taking the limit n → ∞ and qi → dq we have
dq ≤ 0. T
If the cycle is reversed, then the signs of all qi change and we have n
−
i=1
qi ≤ 0. Ti
Thus, for a reversible cycle the equal sign holds. This completes our proof.
1.4.2 Consequences of Clausius’ Theorem B (i) Note first that Eq. (1.45) implies that A dq T is independent of the path joining A and B if the corresponding transformations are reversible. If I and I I are two distinct paths joining A and B we have 0=
dq = T
I
dq − T
II
dq T
··· = I
.... II
24
1 Two Fundamental Laws of Nature
(ii) Next we define the entropy S as follows. Choose an arbitrary fixed state O as reference state. The entropy S (A) of any state A is defined via
A
S (A) ≡
O
dq . T
(1.46)
The path of integration may be any reversible path joining O and A. Thus the value of the entropy depends on the reference state, i.e. it is determined up to an additive constant. The difference in the entropy of two states A and B, however, is completely defined: B dq . S (B) − S (A) = A T Therefore dS =
dq T
(1.47)
for any infinitesimal reversible transformation.
1.4.3 Important Properties of the Entropy (i) For an irreversible transformation from A to B:
B A
dq ≤ S (B) − S (A) . T
(1.48)
Proof: We construct a closed path consisting of the irreversible piece joining A and B and a reversible piece returning to A. Thus 0≥ and therefore
dq = T
irreversible path from A to B
irrev.
dq − T
reversible path from A to B
dq ≤ S (B) − S (A) . T
dq , T
(1.49)
(ii) The entropy of a thermally isolated system never decreases. Proof: Referring to the previous equation thermal isolation means dq = 0. It follows that 0 ≤ S (B) − S (A) or S (A) ≤ S (B) . (1.50) This is the manifestation of the thermodynamic arrow of time. All of the above follows from the two equivalent postulates by Kelvin and Clausius. They constitute the second law of thermodynamics. However, mathematical
1.4 Entropy
25
formulations of the second law are the Clausius theorem or the last two inequalities above. (iii) Another important property of the entropy, as we have shown above, is its sole dependence on the state in which the system is in. Like the internal energy the entropy is a state function whereas q is not (cf. Remark 1 below!). Combining the first law16 with Eq. (1.47) yields dS =
μi 1 P 1 d E + d V − H · d m − dn i + . . . , T T T T
(1.51)
i
where 1 ∂ S = , ∂ E V,m,n,... T ∂ S 1 = P, ∂ V E,m,n,... T ∂ S 1 = − H , E,V,n,... ∂m T and ∂ S 1 = − μi . j=i ... ∂n i E,V,m,n T
(1.52) (1.53) (1.54)
(1.55)
Equation (1.52) may be viewed as a thermodynamic definition of temperature. Note In the also that here the H -field is assumed to be constant and δ m = V d V δ M. If analogous electric case H · d m is replaced by E · d p, where δ p = V d V δ P. the field strengths and the moments are parallel, then we have H · d m = H dm and E · d p = Edp. Notice the correspondence between the pairs (H, m), (E, p) and (P, −V ). In other words, we may convert thermodynamic relations derived for the variables P, −V via replacement into relations for the variables H, m and E, p. Even more V D/(4π general is the mapping (P, −V ) ↔ ( E, )) or (P, −V ) ↔ ( H , V B/(4π )), where we assume homogeneous fields throughout the (constant) volume V . Equation (1.51), including modifications thereof according to the types of work involved during the process of interest, is a very important result! For thermodynamics it is what Newton’s equation of motion is in mechanics or the Schrödinger equation in quantum mechanics—except that here there is no time dependence.17 Remark 1: Thus far we have avoided the special mathematics of thermodynamics, e.g. what is a state function, what is its significance, and why is q not a state
16
The type of work to be included of course depends on the problem at hand. The terms in the following equation represent an example. 17 We return to this point in the chapter on non-equilibrium thermodynamics.
26
1 Two Fundamental Laws of Nature
function etc. At this point it is advisable to study Appendix A, which introduces the mathematical concepts necessary to develop thermodynamics. Remark 2: The discussion of state functions in Appendix A leads to the conclusion that Eq. (1.51) holds irrespective of whether the differential changes are due to a reversible or irreversible process!18
18
We shall clarify the meaning of this in the context of two related equations starting on p. 55.
Chapter 2
Thermodynamic Functions
2.1 Internal Energy and Enthalpy We consider the internal energy to be a function of temperature and volume, i.e. E = E(T, V ). This is sensible, because if we imagine a certain amount of material at a given temperature, T , occupying a volume, V , then this should be sufficient to fix its internal energy. Thus we may write dE =
∂ E ∂ E dT + d V. ∂T V ∂V T
(2.1)
The coefficient of dT is called isochoric heat capacity or heat capacity at constant volume: CV ≡
∂ E . ∂T V
(2.2)
It is useful to define another state function, the enthalpy H , via H = E + P V. We find out on which variables H depends by computing its total differential: d H = d E + d (P V ) ∂ E ∂ E = dT + d V + Pd V + V d P. ∂T V ∂V T Replacing d V via dV =
∂ V ∂ V dT + d P, ∂T P ∂P T
R. Hentschke, Thermodynamics, Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-36711-3_2, © Springer-Verlag Berlin Heidelberg 2014
27
28
2 Thermodynamic Functions
V = V (T, P), leads to ∂ E ∂ E ∂ V ∂ V dH = + +P dT ∂T V ∂V T ∂T P ∂T P ∂ E ∂ V ∂ V + +P + V d P. ∂V T ∂ P T ∂P T Application of Eq. (A.1) with A = E yields ∂ V ∂ E ∂ E ∂ E (. . .) dT = +P + − dT ∂T V ∂T P ∂T P ∂T V ∂(E + P V ) = dT. P ∂T Applying Eq. (A.2) again setting A = E yields ∂ V ∂ E (. . .) d P = +P + V dP ∂P T ∂P T ∂(E + P V ) = d P. T ∂P Thus we find dH =
∂ H ∂ H dT + dP ∂T P ∂P T
(2.3)
and therefore H = H (T, P). Replacing the dependence on volume by a dependence on pressure is of great practical importance. From a theoretical point of view working at fixed volume usually is convenient. But experimenting with a closed apparatus, inside which a process leads to the buildup of uncontrolled pressure, is likely to produce uncomfortable feelings. The coefficient of dT in Eq. (2.3), ∂ H (2.4) CP ≡ , ∂T P is the isobaric heat capacity, i.e. the heat capacity at constant pressure. Two other useful quantities are 1 ∂ V (2.5) αP ≡ , V ∂T P the isobaric thermal expansion coefficient, and 1 ∂ V κT ≡ − , V ∂P T
(2.6)
2.1 Internal Energy and Enthalpy
29
Table 2.1 Selected compounds and values for C P , α P , and κT Compound (20 ◦ C, 1 bar)
Air n-pentane (20 ◦ C, 1 bar) Ethanol (20 ◦ C, 1 bar) Water (20 ◦ C, 1 bar) Water (0 ◦ C, 1 bar) Ice Ih (0 ◦ C, 1 bar) Iron (20 ◦ C, 1 bar)
C P [J/(g K)]
α P [10−4 K−1 ]
κT [10−5 MPa−1 ]
1.007 2.3 2.43 4.18 4.22 2.11 0.45
36.7 16 11 2.06 −0.68 1.59 0.35
106 247 117 45.9 50.9 13.0 0.6
the isothermal compressibility. Selected compounds and values for C P , α P , and κT are listed in Table 2.1. There is no need to discuss these quantities at this point. We shall encounter many examples illustrating their meaning.
2.2 Simple Applications 2.2.1 Ideal Gas Law Here we consider a number of simple examples involving gases. Most of the following applications are based on assuming that the gases are ideal. This means that pressure, P, volume, V , and temperature, T , are related via P V = n RT.
(2.7)
R = 8.31447 m 3 PaK −1 mol −1 .
(2.8)
The quantity R is the gas constant
Figure 2.1 shows P Vmol /R, where P = 105 Pa is the pressure and Vmol is the molar volume of air, plotted versus temperature. The data are taken from HCP (Appendix C). The mass density c in the reference is converted to Vmol via Vmol = m mol /c using the molar mass m mol = 0.029 kg. We note that air at these conditions is indeed quite ideal. Notice also that the line, which is a linear least squares fit to the data (crosses), intersects the axes at the origin. The temperature T = 0 K corresponds to T = −273.15 ◦ C. Remark 1: For an ideal gas we easily work out αP =
1 T
and
κT =
1 . P
(2.9)
30
2 Thermodynamic Functions
Fig. 2.1 The ideal gas law
1000
PVmol /R
800 600 400 200 200
400
600
800
1000
T [K]
Remark 2: Equation (2.7) is simple but nonetheless important, because it is used frequently throughout this text as a first and often satisfactory approximation.
∂E/∂V|T = 0 for an Ideal Gas First we prove that ∂ E =0 ∂V T
(2.10)
for an ideal gas, because we shall rely on this equation a number of ties. Starting from Eq. (1.51) we have T d S = d E + Pd V,
(2.11)
because all other variables like n etc. are constant. Immediately it follows that T
∂ E ∂ S = + P. ∂V T ∂V T
(2.12)
But this does not look like much progress. With some foresight we compute the following differential d(E − T S) = d E − T d S − SdT = −SdT − Pd V.
(2.13)
Thus we find ∂(E − T S) = −S V ∂T
(2.14)
∂(E − T S) = −P. T ∂V
(2.15)
and
2.2 Simple Applications
31
Consequently ∂ ∂(E − T S) ∂ S =− V T ∂V ∂T ∂V T
(2.16)
∂ ∂(E − T S) ∂ P =− . T V ∂T ∂V ∂T V
(2.17)
∂ S ∂ P = . ∂V T ∂T V
(2.18)
as well as
and therefore
For an ideal gas this becomes ∂ S =P. ∂V T
T
(2.19)
Inserting Eq. (2.19) into Eq. (2.12) completes the proof of the above statement. Remark: Integrating Eq. (2.19) using Eq. (2.7) we immediately obtain S(T, V ) − S(T, Vo ) = n R ln
V . Vo
(2.20)
This means that if an ideal gas is compressed (expanded) isothermally, i.e. V < Vo (V > Vo ), its entropy is decreased (increased).
Kinetic Pressure Concrete thermodynamical calculations require concrete models. Here we consider a gas of point particles with masses m, i.e. atoms or molecules without internal structure or specific spatial extend and shape, confined to a volume V . The particles posses a momentum distribution d N p = N f (| p|)d 3 p. N is the total particle number and d N p is the fraction of particles whose momenta occupy a momentum space element d 3 p. The quantity f (| p|) is the attendant momentum probability density. Figure 2.2 shows a particle reflected by one of its containment’s walls. The sum of very many (simultaneous) momentum transfers, pz = 2mz/t, each contributing a force f z = pz /t, yields the pressure P = F/A, where F is the total force and A is the wall area. • kinetic pressure: First we want to show that 1 P= V
d N p ( p · n) v p · n ,
(2.21)
32
2 Thermodynamic Functions
Fig. 2.2 A particle reflected from a wall
n
z
z
where p = m v p is the momentum of a particle with velocity v p . The prime is a reminder that the angle θ is between 0 and π/2 for particles impinging on the wall. We consider a single particle for which fz =
2m| v p · n| pz = . t t
Here z is the thickness of a narrow layer adjacent to the wall in which the momentum transfer occurs. This layer is ill defined, because we consider point particles interacting with a completely smooth wall. Fortunately the final result does not require to specify z assumed to be the same for all particles. We define the collision time, i.e. the time a particle spends inside the layer, t, via v p,z = 2z/t, i.e. | v p · n| 1 = . t 2z Combination of the two formulas yields fz =
( p · n)( v p · n) . z
(2.22)
In order to obtain the total force on the wall exerted by the gas we must sum or integrate over all collisions, i.e. F=
z A d N p fz V
(2.23)
∗
*: number of particles inside the surface layer volume z A possessing momenta p in d 3 p. Limiting θ to values between zero and π/2 includes particles colliding with the wall only. Thus we obtain Eq. (2.21).
2.2 Simple Applications
33
• ordinary particles: For an ideal gas1 we can deduce the important result 3 n RT. 2
E=
(2.24)
Consequently the isochoric heat capacity of an ideal gas of point particles is given by 3 n R. 2
CV =
(2.25)
In order to show Eq. (2.24) we express the internal energy of the gas via
E=
d N p
p2 . 2m
(2.26)
Notice that we continue to use the prime (consistent with the normalization of our above probability density f (| p|)). It is now easy to deduce P=
2E . 3V
(2.27)
This is because the θ -integration in the case of E is
π/2
π/2
dθ sin θ = −
0
d cos θ =
0
1
dx = 1
0
and thus
2π
E=
dφ 0
∞
−∞
dpp 2
p2 f ( p). 2m
(2.28)
In the case of P we have instead
π/2
cos θ sin θ dθ = 2
0
1
x 2d x =
0
1 . 3
and therefore P=
1 3V
2π
dφ 0
∞
−∞
dpp 2
p2 f ( p). m
(2.29)
Comparison of Eqs. (2.28) and (2.29) immediately yields Eq. (2.27), which in combination with the ideal gas law yields Eq. (2.24). 1 Whether or not our assumption of point particles already implies ideality is a matter of definition of the inter particle interactions.
34
2 Thermodynamic Functions
Finally we take another look at the entropy. We have dS =
P 1 d E + dV T T
(2.30)
cf. (1.51). Using Eq. (2.24), i.e. d E = (3/2)n RdT , and once again Eq. (2.7) we find
d S = n R d ln T 3/2 − d ln V .
(2.31)
Integration yields the generalization of Eq. (2.20) S(T, V ) − S(To , Vo ) = n R ln
T To
3/2
V . Vo
(2.32)
This is the ideal (point particle) gas entropy change as function of temperature and volume. • photons: Now let us assume that the gas particles are photons obeying the energymomentum relation = cp. Inserting this relation into Eqs. (2.21) and (2.26) yields 1 P= V
1 d N p pc cos θ = 3V 2
d N p pc
(2.33)
and E=
d N p pc.
(2.34)
Notice that v p is replaced by the velocity of light, c. Comparison of the two equations produces P=
1E . 3V
(2.35)
We can even relate the photon pressure to the gas temperature via ∂ E ∂ P =T −P ∂V T ∂T V
(2.36)
derived in the previous example. Because the classical energy, E, of the photon gas is given by 1 E 2 + H 2 E = , (2.37) dV V V V 8π where the argument of the integral is the electromagnetic energy density, we conclude that P does not depend on V and therefore
2.2 Simple Applications
35
∂ E = 3P. ∂V T The resulting differential equation is 4
dT dP = or d ln T 4 = d ln P T P
and thus P = Po
T To
4 (2.38)
or E = cr T 4 , V
(2.39)
where cr is a constant. This is Stefan’s law of the energy density dependence on temperature in the case of black body radiation. This also is a good place to mention the entropy of the black body, i.e. a volume containing (and possibly emitting) radiation in thermal equilibrium with some reservoir.2 We start by inserting Eq. (2.39) into the thermodynamic definition of temperature, Eq. (1.52), i.e. 1 ∂S = . 3 4cr V T ∂ T V,n,... T
(2.40)
Integration of this equation yields the entropy of the black body S=
4E , 3T
(2.41)
which is proportional to T 3 . Notice that S = 0 at T = 0. However, there is more to learn here. Subtracting the two differentials 4 T d S = 4cr V T 3 dT + cr T 4 d V and d E = 4cr V T 3 dT + cr T 4 d V 3
(2.42)
yields TdS − dE =
1 (2.35) cr T 4 d V = Pd V. 3
(2.43)
2 The term “black body” may be somewhat misleading, because a black body is not necessarily black. In fact the radiation spectrum of our sun measured above the atmosphere is very closely a black body spectrum. Here we merely deal with the temperature dependence of the total energy density of a black body. The spectrum is calculated in Sect. 5.2.
36
2 Thermodynamic Functions
Fig. 2.3 Experimental setup in the author’s office allowing to verify Stefan’s law 6 5
[mV]
4 3 2 1 5
10
15
T 4 _ To4
20
25
[K 4]
1010 Fig. 2.4 Data obtained with the experimental setup (including cooling of the aperture plate) shown in the previous figure
Comparing this result to Eq. (1.51) we conclude that the chemical potential of the photons vanishes, μ = 0. We shall return to this conclusion later in this book (on page 213). Figure 2.3 shows an experimental setup in the author’s office allowing to verify Stefan’s law. The red cube in the center is an oven (black body) emitting radiation through the aperture shown in the inset. The radiation energy is measured and converted into volts shown on the instrument panel on the right. The attendant oven temperature is shown by the instrument on the left. Fig 2.4 contains data taken by the author upon heating (up-triangles) and subsequent cooling (down-triangles) of the oven (To = 25 ◦ C). The solid line is a linear fit through the data points. Even though there is some room for improvement the result is clearly in accord with Eq. (2.39).
2.2 Simple Applications
37
It is interesting to calculate the energy a black body looses per unit time due to radiation emanating from its surface. If δ A is an area element on the black body’s surface, a distant observer may look at δ A from an angle θ . Here θ is the angle between the surface normal of δ A and the direction of the observer. Thus the observer does not see the full δ A but the projection δ A cos θ instead. Every volume element on the surface contains the energy density E/V , and therefore emanates a total flux density cE/V . Along a particular direction this is cE/(4π V ). This means that the energy per time passing through the area element projection in the direction of the observer is cE/(4π V )δ A cos θ . Collecting together the energy per time passing through δ A towards all possible observer directions therefore yields cE dδ E =− δA dt 4π V
2π
π/2
dϕ 0
dθ sin θ cos θ = −
0
1 cE δA 4 V
(2.44)
or after integration over the full surface 1 cE dE =− A = −σ T 4 A. dt 4 V
(2.45)
The −-sign indicates that the energy of the black body diminishes. The quantity σ = 5.67 · 10−8 Wm−2 K−4 is Stefan’s constant, which we learn how to calculate in Sect. 6.2, cf. (6.102).3 Remark 1: Black Hole Entropy. In the early 1970s Stephen Hawking (Hawking 1976; and references therein) showed that a black hole of mass M should emit radiation possessing the spectral distribution of black body radiation4 corresponding to the temperature Tbh =
1 c3 . 8π k B G M
(2.46)
Here k B is the so-called Boltzmann constant, G is the gravitational constant, is Planck’s constant divided by 2π , and c is the speed of light. The origin of this radiation is the separation of virtual particle pairs, spontaneously created just outside the event horizon of the black hole due to vacuum fluctuations, such that the partner with positive energy is traveling away from the horizon and the other in turn vanishes behind the horizon. This negative energy partner diminishes the energy of the black hole by an amount that is carried away by the other. In this sense a black hole may 3 It is interesting to apply this formula to the sun. We use a solar surface temperature of 5700 K, a sun radius of r S ≈ 7.0 · 108 m, an earth radius of r E ≈ 6.4 · 106 m, and the mean sun-to-earth distance R S E ≈ 1.5 · 1011 m. With these numbers we calculate a radiation energy annually received by the earth’s surface of about 1.5 · 1015 MWh/y. At the time of this writing the entire world’s electricity consumption is roughly 1.9 · 1010 MWh/y (based on data collected between 2002 and 2010). In other words—a quadratic surface of about 40 by 40 km positioned in space near the earth would receive just this energy! 4 This distribution is shown for one particular temperature in Fig. 5.12.
38
2 Thermodynamic Functions
“evaporate” over time. As already mentioned, remarkably this radiation has the same distribution as the radiation of a black body—despite its different nature! According to Eq. (1.52) the black hole should posses entropy, which we can calE culate by integrating this equation (S = 0 d E T (E )−1 + const): Sbh = 4π
kB G M2 . c
(2.47)
We have used E = Mc2 (d E = c2 d M) and assumed that const = 0. Notice that Eq. (1.52) holds if the other variables (V , . . .) are held fixed. In the case of the black hole the corresponding quantities are its charge and angular momentum. The entropy in Eq. (2.47) may be cast in a different form, i.e. Sbh =
kB A . 4 λ2P
(2.48)
A = 16π G 2 M 2 /c4 is the horizon area5 and λ2P the square of the Planck length λ P = (G/c3 )1/2 ∼ 10−35 m. Equation (2.48) is the Bekenstein-Hawking entropy formula. Jacob Bekenstein (Bekenstein 1973) was the first to systematically explore that “a black hole exhibits a remarkable tendency to increase its horizon surface area when undergoing any transformation” in analogy to the second law of thermodynamics (as expressed via Eq. (1.50)). He concluded that a black hole should posses entropy proportional to A. His reasoning, predating the discovery of black hole radiation, was based on the connection between entropy and information or rather the lack of information.6 The formula for Sbh , which he obtained, differs from Eq. (2.48) by a numerical factor (he introduced λ2P on dimensional grounds!). We may estimate the time tvap (very roughly!) it should take for the black hole to evaporate via −
dE 1 dM ∼− ∼ T4A ∼ 2, dt dt M
(2.49)
i.e.
tvap 0
dt ∼
M
d M M . 2
(2.50)
0
Our final result is
5 This formula may be obtained classically! The velocity necessary to escape from a mass
M starting at a distance R is vesc = (2G M/Rbh )1/2 . Substituting vesc = c and solving for Rbh yields A via 2 . A = 4π Rbh 6 This connection is discussed on p. 185
2.2 Simple Applications
39
Table 2.2 Characteristic quantities for black holes with selected masses Sun Earth
M (kg)
Rbh (m)
Tbh (K)
tvap (s)
2 · 1030 6 · 1024 4 · 1022 7 · 1011 1
3 · 103 9 · 10−3 6 · 10−5 10−15 10−27
6 · 10−7 6 · 10−2 2.75 2 · 1011 1023
1070 1054 1047 1015 10−20
tvap ∼
G2 M 3 , c4
(2.51)
where the factor G 2 /(c4 ) follows from dimensional analysis. Table 2.2 compiles some numbers for M equal to the mass of the sun, the earth, and two arbitrary masses (one corresponding to the cosmic background temperature discussed in the context of Fig. 6.12), and illustrates the extreme numbers (The age of the universe is estimated at 5 · 1017 s!). This shows that the radiation is significant only for black holes much smaller than those that are expected to form by the collapse of stars. Black holes possessing smaller masses could have formed in the early universe, but thus far no conclusive experimental evidence for black body radiation, especially for the brief intense flash indicating a vanishing black hole, has been found. A detailed discussion of the underlying theory is given in Susskind and Lindsay (2005). Remark 2: Temperature in an Expanding Universe. We live in an expanding universe believed to have no exchange with an “outside”. Thus δq = 0 and according to Eq. (1.1) we conclude that energy and volume of the universe are related via d E = −Pd V
(2.52)
d(εr V + E m ) = −(Pr + Pm )d V.
(2.53)
or
Here εr = cr T 4 is the radiation energy density according to Eq. (2.39) and cr is a constant.7 The quantity E m we approximate via E m = (3/2)n RT + n N A mc2 . The first term is the ideal gas contribution according to Eq. (2.24) and the second term is the rest energy of the gas of particles (m: rest mass; c: speed of light). This is we assume a universe containing two kinds of matter, one is electromagnetic radiation and the other is a constant number of particles like neutrons and protons. The relation between pressure and energy density for the two types of matter is given by the Eqs. (2.35) and (2.27). Inserting this into Eq. (2.53) yields
7
This constant is derived in Chap. 5; cf. (5.102).
40
or
2 Thermodynamic Functions
3 1 n RT cr T 4 + dV d(cr T 4 V + n RT + n N A mc2 ) = − 2 3 V
(2.54)
3 n RT 4 3 4 4cr T V + n R dT = − d V. cr T + 2 3 V
(2.55)
After some algebra we find Y
nR cr V T 3
d ln T = d ln Ru−1 ,
where Y (x) =
1 + 3x/8 , 1 + 3x/4
(2.56)
(2.57)
and Ru ∝ V 1/3 is the radius of the universe. In the limit x → 0 the radiation dominates and Y = 1. In the limit x → ∞ the particles dominate and Y = 1/2. According to Eq. (2.56) we find the following limiting relations between the temperature of the universe and its radius: −1 Ru Y = 1 radiation dominates . (2.58) T ∝ Ru−2 Y = 2 particles dominate In both cases does a lowering of the temperature occur as the universe expands. Below a certain radius the particles dominate this relation and the temperature drops more quickly in terms of Ru . Above this radius the radiation dominates and the temperature drops less rapid. For those who are interested in a deeper discussion of the implications we recommend Zhi and Xian (1989).
Isotherms and Adiabatic Curves Discussing the Carnot engine we had studied the thermodynamic cycle in the P-V plane depicted in Fig. 2.5. The two curves labeled T1 and T2 are called isotherms (T = constant). The two other curves are adiabatic curves (δq = 0). Starting from the ideal gas law, P V = n RT , we may show that the sketch is correct in so far as the isotherms are less steep than the adiabatic curves, i.e. ∂ P ∂ P . > ∂V T ∂ V δq=0 Notice that the combination of the first law, i.e. d E = −Pd V (if δq = 0), with Eq. (2.2) yields (2.59) − Pd V = C V dT.
2.2 Simple Applications
41
Fig. 2.5 Carnot cycle
P a q=0
q2 b
T1
T2 d q1
q=0 c
V
From dT =
∂ T ∂ T dV + dP ∂V P ∂P V
follows dT =
V P dV + d P. nR nR
This means −Pd V = C V
P V dV + dP nR nR
or
CV − 1+ nR
dV CV d P = V nR P
and therefore d ln P nR = −1 − . d ln V δq=0 CV
(2.60)
Along an isotherm we have d P n RT P =− 2 =− dV T V V and therefore d ln P = −1. d ln V T Combination of Eqs. (2.60) and (2.61) yields
(2.61)
42
2 Thermodynamic Functions
d ln P d ln P > d ln V T d ln V δq=0 or d P d P . > dV T d V δq=0
Efficiency of Engines with Ideal Gas as Working Substance We study three examples: (a) the Carnot engine or cycles, (b) the Otto cycle, and (c) the Diesel cycle. (a) Figures 1.10 and 2.5 both show the Carnot cycle. Assuming that the working medium is an ideal gas we want to compute the thermal efficiency for this cycle. We consider the work done by the gas along the different parts of the cycle: a→b:
wa→b =
b→c: c→d:
wb→c = wc→d = wd→a =
Va Vc
Vb Vd
d→a:
Vb
Vc Va
δT =0
Pd V = n RT2 ln δq=0
Vb Va
T1
Pd V = −C V
dT = −C V (T1 − T2 )
T2 δT =0
Pd V = n RT1 ln δq=0
T2
Pd V = −C V
Vd
Vd Vc dT = −C V (T2 − T1 ).
T1
The total work done by the gas is w = wa→b + wb→c + wc→d + wd→a = wa→b + wc→d .
(2.62)
Now we compute the heat input q2 . Notice that for an ideal gas, as we just have seen, E = E(T ). The path from a to b is along an isotherm however and therefore 0 = E a→b = q2 −wa→b , i.e. q2 = wa→b . We thus obtain for the thermal efficiency η=
wc→d T1 ln[Vc /Vd ] w . =1+ =1− q2 wa→b T2 ln[Vb /Va ]
(2.63)
Integrating Eq. (2.59) for the ideal gas yields V = V
T T
C V /(n R) (2.64)
along an adiabatic curve. We can use this to express Vd via Va and Vc via Vb . The result is
2.2 Simple Applications
43
Fig. 2.6 Otto cycle
P
c qb
wc
d
c
b q=0
d
q=0
a
wa
η =1−
qd
a
V b
T1 T2
(2.65)
in accord with Eq. (1.44). (b) Figure 2.6 shows the Otto cycle. The contributions from the different parts of the cycle are a → b : qa→b = 0 − wa→b = C V (Tb − Ta ) b → c : wb→c = 0 qb→c = C V (Tc − Tb ) c → d : qc→d = 0 − wc→d = C V (Td − Tc ) d → a : wd→a = 0 qd→a = C V (Ta − Td ). The thermal efficiency is w qb→c
=
Td − Ta −C V (Tb − Ta ) − C V (Td − Tc ) =1− . C V (Tc − Tb ) Tc − Tb
Along the adiabatic curves we have Ta = Tb
Vb Va
(n R)/C V and
Td = Tc
Vc Vd
(n R)/C V
.
With Vb = Vc and Va = Vd follows Ta Td = Tb Tc
or
and therefore η =1−
Td =
Ta . Tb
Ta Tc Tb
(2.66)
44
2 Thermodynamic Functions
Fig. 2.7 Diesel cycle
P
qb
c
wb
b
c
wc
c
d
q=0
q=0
d a
wa
qd
a
V b
(c) Figure 2.7 shows the Diesel cycle. This cycle is similar to the Otto cycle. The difference is that the isochor, i.e. the line of constant volume from b to c, is changed to an isobar, a line of constant pressure. The contributions from the different parts of the cycle are in this case a → b : qa→b = 0
− wa→b = C V (Tb − Ta )
b → c : wb→c = Pb (Vc − Vb ) qb→c = C V (Tc − Tb ) + wb→c c → d : qc→d = 0 − wc→d = C V (Td − Tc ) d → a : wd→a = 0 qd→a = C V (Ta − Td ). Here the thermal efficiency is w −C V (Tb − Ta ) + Pb (Vc − Vb ) − C V (Td − Tc ) . = qb→c C V (Tc − Tb ) + Pb (Vc − Vb ) Using the ideal gas law together with Eq. (2.64) this may be rewritten as η =1−
Vb Va
n R/C V
1 (Vc /Vb )1+n R/C V − 1 . 1 + n R/C V Vc /Vb − 1
(2.67)
Notice that there is the following relation between the efficiencies of Otto and Diesel cycles: 1−η =
Vb Va
γ −1
g
(2.68)
Diesel Otto
(2.69)
where g=
1 x γ −1 γ x−1
1
2.2 Simple Applications
T
45
T
2
2
II
3
I 3
1
a
b
1
a
S
b S
Fig. 2.8 Two cycles in the T-S-plane
with x = Vc /Vb and γ = 1 + n R/C V . One can show (exercise8 ) that g Diesel > 1 for x > 1 and γ > 1. This means that a reversible Otto cycle with an ideal gas as working substance is more efficient than a reversible Diesel cycle at the same compression ratio, Vb /Va . However, real Diesel engines can operate at greater compression ratios and therefore greater efficiencies than Otto engines. Remark: In Fig. 1.11we compare the Carnot engine to a competing X -engine. A conclusion in the attendant discussion is that ηCar not = η X if X is reversible. This may inspire the idea to replace X with reversible Diesel and Otto engines. The result is that both engines should have the same efficiency in contradiction to the above calculation. Where is the mistake? If we apply Eq. (1.37), i.e. η = 1 − q1 /q2 , to our two engines using q1 = −qd→a and q2 = qb→c , then we can indeed use the same η = 1 − (−qd→a )/qb→c in both cases. The respective results agree with the results of (b) and (c). The difference is that the two cycles are not the same, leading to different efficiencies for identical compression ratios.
Cycles in the T -S-Plane Figure 2.8 shows two reversible cycles in the T -S-plane (scales are identical). Which of the two has the greater thermal efficiency? The basic equation is d E = T d S − dw.
(2.70)
In both cases E = E(T, S) and cycle d E = 0, because E is a state function. The work done during one complete cycle is
dw =
w= cycle
TdS
(2.71)
cycle
(clockwise). The heat absorbed is obtained by integration along the parts of the cycle for which d S > 0, i.e. Idea: (a) expansion in terms of x − 1 near x = 1 shows that g > 1 in this limit; (b) comparison of x-derivatives of denominator and numerator of g.
8
46
2 Thermodynamic Functions
qin =
dw = cycle,>
T d S.
(2.72)
cycle,>
Thus the two thermal efficiencies are η=
area 1-2-3-1 w . = qin area a-2-3-b-a
(2.73)
While the areas in the numerators are equal, the area a-2-3-b-a is larger for cycle II and therefore ηI > ηI I .
(2.74)
Temperature Profile of the Troposphere At high altitude the air temperature may be much lower than the ground temperature as some of us know from traveling on airplanes. How can we explain this? We consider an air bubble rising in the atmosphere. According to the first law the differential change of its internal energy is d E = δq − Pd V = C V dT +
∂ E d V. ∂V T
Assuming that the air is an ideal gas the last term on the right is zero as we have just shown. We also assume that the bubble does not exchange heat and thus rises adiabatically, i.e. δq = 0. Again we simply have − Pd V = C V dT.
(2.75)
We want the temperature, T , expressed in terms of the height, h, the air bubble has risen. The quantity which we may connect easily to h is the pressure—as we shall see. Therefore we use the ideal gas law to replace d V via − Pd V = −n RdT + n RT d ln P.
(2.76)
Combination of Eqs. (2.75) and (2.76) yields zd ln T = d ln P,
(2.77)
where z = C V /(n R) + 1. In order to express P in terms of h we consider a column of air parallel to the gravitational field of the earth as shown in Fig. 2.9. The pressure at the bottom of the column element (solid cube), P(h), is related to the pressure at its top, P(h + δh), via
2.2 Simple Applications
47
Fig. 2.9 A column of air parallel to the gravitational field
h
h
A mair
h
P(h) = P(h + δh) +
δm air g . A
(2.78)
Here δm air is the mass of the air contained in the column element and g is the gravitational acceleration. Via the expansion P(h + δh) ≈ P(h) + (d P(h)/dh)δh we find d P(h) = −cgdh,
(2.79)
where c = δm air /(Aδh) is the mass density in the column element. c of course depends on h, the position of the element in the column. Assuming that the column element contains n moles of air we write c = nm mol /(n RT /P) using the ideal gas law. Here m mol ≈ 0.21m O2 + 0.78m N2 ≈ 0.029 kg is the molar mass of air, where m O2 ≈ 0.032 kg and m N2 ≈ 0.028 kg are the molar masses of oxygen and nitrogen. Thus Eq. (2.79) turns into d ln P = −
To dh , T Ho
(2.80)
with Ho =
RTo , m mol g
(2.81)
i.e. Ho ≈ 29.2To mK −1 , where To is the air temperature at h = 0. Usually Ho is close to 8500 m. Combination of Eq. (2.77) with Eq. (2.80) yields after integration h . T = T0 1 − z Ho
(2.82)
48
2 Thermodynamic Functions
According to this equation the temperarture drops linearly with increasing altitude. However, we need z before we can compute concrete numbers. We may look up z in a table. In Ref. HCP we find C P = 1.007 JK −1 g−1 at T = 300 K and P = 105 Pa. For an ideal gas C P = C V + n R (cf. p. 65) and therefore z ≈ 3.5.9 Thus, according to Eq. (2.82) the temperature reaches absolute zero at around 3 · 104 m. Before we compare this to the experimental data we want a corresponding pressure profile P(h), which is readily obtained by inserting Eq. (2.82) into Eq. (2.80). We find d ln P = −z
dh . z Ho − h
(2.83)
If z Ho h we can neglect h in the denominator and the pressure profile becomes P = P0 exp [−h/Ho ] ,
(2.84)
where P0 is the pressure at h = 0. Equation (2.84) is called barometric formula. Integration of the full Eq. (2.83) yields P = P0 (T /To )z
(2.85)
instead. Figure 2.10 summarizes our results (solid lines—Eqs. (2.82) and (2.85); dashed line—Eq. (2.84)) and compares them to data (crosses) taken from the literature (here: “http://www.usatoday.com/weather/wstdatmo.htm”; “Source: Aerodynamics for Naval Aviators”). Notice that the temperature data are not direct measurements but rather data points computed from simple formulas describing the average temperature profile at different heights. Our calculation applies to the troposphere, i.e. to a maximum altitude of roughly 10000 m. Beyond the troposphere other processes determine the temperature of the atmosphere. We see that our result somewhat underestimates the actual temperature. The middle graph shows the pressure profile. We notice that the two theoretical models Eq. (2.84) (isothermal case10 ) and Eq. (2.85) (adiabatic case) bracket the true pressure profile. The bottom graph shows the compressibility factor, Z = P V /(n RT ), versus h. The data points scatter because of scatter in the density values. Nevertheless the graph shows that our assumption of ideal gas behavior is very reasonable.
Some of you may already know that C V /(n R) = 3.5 = 7/2 on the basis of the so called equipartition theorem, because every degree of freedom contributes 1/2 to C V /(n R). Every O2 - and every N2 -molecule, the majority of what air consists of, has three center of mass kinetic degrees of freedom (3 · 1/2; cf. (2.25)). In addition both have two axes of rotation (2 · 1/2). Finally they both are one-dimensional oscillators (2 · 1/2). A more detailed, i.e. quantum theoretical, calculation reveals that these two degrees of freedom do not contribute at the temperatures considered here. Therefore C V /(n R) = 5/2 to good approximation and thus C P /(n R) ≈ 7/2. 10 With T = T we can directly integrate Eq. (2.80) to obtain the barometric equation. o 9
2.2 Simple Applications
49
T [ oC ] 10 5
10
15
20
25
30
35
5
10
15
20
25
30
35
5
10
15
20
25
30
35
h [103 m ]
-10 -20 -30 -40 -50
P [105 Pa ] 1.0 0.8 0.6 0.4 0.2
h [103 m]
Z 1.10 1.05 1.00 0.95
0
h [103 m]
Fig. 2.10 Summary of results
Before moving on we want to estimate one interesting number—the total mass of the earth’s atmosphere, Matm . Notice that the ground pressure is Po = Matm g/ (4π R 2E ), where R E ≈ 6.37 · 106 m is the earth’s radius. With Po = 1 bar the total mass of the atmosphere is Matm ≈ 5.2 · 1018 kg.
50
2 Thermodynamic Functions
Fig. 2.11 A volume element experiencing a pressure difference
P
P
u
P
A
x
r
Speed of Sound in Gases and Liquids Figure 2.11 depicts a volume element in a medium. The medium can be a gas or a liquid. The volume element experiences a pressure difference along the x-direction. This means that the left face of the element experiences the pressure P while the right face is under slightly higher pressure P +δ P. Here P = P¯ is a constant average pressure in the medium, whereas δ P = δ P( r , t) depends on position and time. In order to derive an expression for the speed of sound we work from the continuity equation ∂ ( c( r , t) + ∇ u ( r , t)c( r , t)) = 0. ∂t
(2.86)
Here c( r , t) is the mass density inside the volume element. Again we assume c( r , t) = c¯ + δc( r , t). The pressure modulation causes a corresponding slight spatial and temporal variation of the density relative to its average, c. ¯ The quantity u( r , t) is the instantaneous velocity of the volume element due to the pressure gradient. Using the approximation u( r , t)c( r , t) ≈ u( r , t)c¯ and taking another partial derivative with respect to time yields ∂2 ∂ r , t) ≈ 0. δc( r , t) + ∇ c¯ u( ∂t 2 ∂t
(2.87)
The term in brackets is the average mass density times the acceleration of the volume element, which can be expressed via the pressure gradient according to the equation P( of motion c∂ ¯ u( r , t)/∂t = −∇ r , t). Therefore Eq. (2.87) becomes ∂2 2 δ P( δc( r , t) − ∇ r , t) ≈ 0. ∂t 2
(2.88)
In a final step we express δ P( r , t) in terms of δc( r , t) using the thermodynamic definition of the adiabatic compressibility,
2.2 Simple Applications
51
κS = −
1 ∂ V , V ∂P S
(2.89)
defined analogously to the isothermal compressibility in Eq. (2.6). Adiabatic in the present context means that the density changes in the volume element are fast and no heat is transferred during the fluctuation. On p. 66 we work out in detail the relation between κ S and κT . But for the moment we make use of −δV /V = δc/c and obtain δ P ≈ (cκ ¯ S )−1 δc. This immediately yields the (density) wave equation ∂2 1 2 δc( r , t) − r , t) ≈ 0. ∇ δc( ∂t 2 cκ ¯ S
(2.90)
The velocity of the waves, the sound waves, is vs = √
1 . cκ ¯ S
(2.91)
First we want to apply this formula to air and we ask: What is the speed of sound in this medium? Air is an ideal gas for our purpose. When we work out κ S in detail (beginning on p. 66) we also show that in an ideal gas κ S = (z−1)/(z P). The quantity z = C V /(n R) + 1 was introduced in the context of Eq. (2.77). As in the previous example, the temperature profile of the troposphere, we use C V /(n R) = 5/2. Thus in air 7RT , (2.92) vs = 5m mol where m mol ≈ 0.029 kg is the air’s molar mass. At T = −100 ◦ C we calculate vs = 264 m/s, while for T = 25 ◦ C we find vs = 346 m/s. Both numbers are in excellent agreement with vs -values tabulated in HCP. And what is the speed of sound in water? Here we need to know κ S . Below we shall show that κ S = κT − T V α 2P /C P , cf. (2.153). In Table 2.1 we find the necessary values for water at for instance 20 ◦ C. We also find that κ S ≈ κT in this case, and we obtain vs = 1481 m/s at this temperature—again in very good agreement with the corresponding value in HCP.
Joule-Thomson Coefficient The release of propane gas from its metal can causes significant cooling of the latter. However, there may be situations when a gas leak causes heating and the possible danger of explosion. The so called Joule-Thomson coefficient, μJ T =
∂ T , ∂P H
(2.93)
52
2 Thermodynamic Functions
throttle
P1, V1
P2, V2
F2
F1
Fig. 2.12 A gas being pushed through a throttle
is the quantity which tells us whether the temperature will increase or decrease in such a process. Here μ J T > 0 means cooling (refrigerator) whereas μ J T < 0 means heating. The general process is depicted in Fig. 2.12. A gas initially is under pressure P1 and confined to a volume V1 . In an adiabatic process (δq = 0) the gas is pushed through a throttle and expands into the volume V2 , where the pressure is P2 . According to the first law the internal energy change is E = E 2 − E 1 = w. The net amount of work is w = P1 V1 − P2 V2 , because P1 V1 is the work done to the system and −P2 V2 is the work done by the system. Overall we find E 1 + P1 V1 = E 1 + P1 V1 and therefore H = 0. The process is said to be isenthalpic. This is why the derivative in Eq. (2.93) is at constant enthalpy. For concrete computations Eq. (2.93) may be transformed into μJ T =
1 CP
T
∂ V −V ∂T P
(2.94)
or μJ T
TV = CP
1 αP − . T
(2.95)
Let us find out how to get from Eqs. (2.93) to (2.94). We start with μJ T =
∂ T (A.3) ∂ T ∂ H = − ∂P H ∂H P ∂P T 1 ∂H =− . CP ∂ P T
(2.96)
The quantity ∂ H/∂ P|T is similar to the quantity ∂ E/∂ V |T calculated on p. 30, i.e. the general approach is analogous. Via Eq. (2.11) it follows immediately that TdS = dH − VdP
2.2 Simple Applications
53
and thus T
∂ H ∂ S = − V. ∂P T ∂P T
(2.97)
Similar to the derivation of ∂ E/∂ V |T we now compute d(H − T S) = d H − T d S − SdT = −SdT + V d P.
(2.98)
Here we find ∂(H − T S) = −S P ∂T
(2.99)
∂(H − T S) = V. T ∂P
(2.100)
and
Using the interchangeability of partial derivatives, i.e. ∂ ∂(H − T S) ∂ S =− P T ∂P ∂T ∂P T
(2.101)
∂ ∂(H − T S) ∂ V = , T P ∂T ∂P ∂T P
(2.102)
∂ S ∂ V =− . ∂P T ∂T P
(2.103)
and
we finally obtain
Combination of this equation with Eqs. (2.96) and (2.97) completes the proof. If we assume that V ∝ T k then the Joule-Thomson coefficient becomes μJ T ∝
1 1 kT T k−1 − T k = (k − 1) T k . CP CP
Because C P > 0 (we shall show this), we find that for the ideal gas μ J T = 0. If the gas is not ideal we have μ J T > 0 for k > 1, which means cooling, and μ J T < 0 for k < 1, which means heating. We can now go ahead and measure k in order to find out what will happen. Below we return to the Joule-Thomson coefficient in the context of the van der Waals theory. The latter provides insight as to why a gas will do one or the other.
54
2 Thermodynamic Functions
2.3 Free Energy and Free Enthalpy In the preceding section we found the two quantities E − T S, cf. (2.13) and H − T S, cf. (2.98) to be rather useful. We therefore define the two new functions called the free energy11 F = E −TS
(2.104)
G = H − T S.
(2.105)
and the free enthalpy12
Computing their total differentials we find d F = d E − d(T S) = d E − SdT − T d S (1.51)
= −SdT − Pd V + μdn + . . .
(2.106)
and dG = d H − d(T S) = d H − SdT − T d S (1.51)
= −SdT + V d P + μdn + . . . ,
(2.107)
where ∂ F = −S, ∂ T V,n,...
(2.108)
∂ F = −P, ∂ V T,n,...
(2.109)
∂ F = μ, ∂n T,V,...
(2.110)
∂G = −S, ∂ T P,n,...
(2.111)
and analogously
11 or also Helmholtz free energy; Hermann Ludwig Ferdinand von Helmholtz, german physiologist and physicist, * 31.8.1821 Potsdam, Germany; †8.9.1894 Charlottenburg, Germany. 12 or also Gibbs free energy; Josiah Willard Gibbs, american scientist, * 11. 2.1839 New Haven, Connecticut, †28.4.1903 New Haven, Connecticut; the founder of modern thermodynamics.
2.3 Free Energy and Free Enthalpy
55
∂G = V, ∂ P T,n,...
(2.112)
∂G = μ. ∂n T,P,...
(2.113)
Obviously F = F(T, V, n, . . . ) whereas G = G(T, P, n, . . . ). The two are related via G = F + P V,
(2.114)
i.e. they are Legendre transformsof each other (G = F − V ∂ F/∂ V |T and F = G − P∂G/∂ P|T ). Remark 1: F is called a thermodynamic potential with respect to the variables T , V , n, . . . . The same is true for G with respect to T , P, n, . . . . In general we call a thermodynamic quantity a thermodynamic potential if all other thermodynamic quantities can be derived from partial derivatives with respect to its variables. Remark 2: By straightforward differentiation and knowing that S and E are state functions it is easy to show that F and G are state functions also.
2.3.1 Relation to the Second Law According to the first law we have d E = δq − δw.
(2.115)
Combination of this with Clausius’ statement of the second law, cf. (1.48) yields d E − T d S ≤ −δw
(2.116)
d F |T +δw ≤ 0.
(2.117)
or
If δw stands for volume work only, then we may deduce d F |T,V ≤ 0.
(2.118)
From this follows (cf. p. 286) the attendant relation for the free enthalpy, i.e. dG |T,P ≤ 0.
(2.119)
56
2 Thermodynamic Functions
Fig. 2.13 An illustration of the relation of G to the second law
G dG|
<0
T,P
dG =-SdT + VdP
P T
An illustration in the case of the free enthalpy is shown in Fig. 2.13. A system initially may be prepared in a state corresponding to the solid circle. It lowers its free enthalpy as much as possible, which brings it down to a point on the surface shown in the sketch, where dG is given by Eq. (2.107). This surface represents the states of lowest free enthalpy above the T -P-plane. Of course there may be other types of work, not just volume work, involving variables, X , other than V , which are controlled from outside the system (the examples below explain what is meant here). In this case the Eqs. (2.118) and (2.13) still apply if . . . |T,V and . . . |T,P are replaced by . . . |T,V,X and . . . |T,P,X . This book contains a number of applications of both Eq. (2.118), e.g. phase separation in the context of van der Waals theory, and Eq. (2.119), e.g. chemical reactions. Nevertheless, already at this point we want to look at three instructive examples. Example: Capillary Rise. Figure 2.14 shows a tube of radius R with one end submerged in a liquid. In the figure the liquid has risen to a height h against the pull of gravity. We want to calculate the equilibrium height of the liquid in the tube. This system comprises the liquid, the tube, and the earth. Ignoring all other effects we use δ F |T = (δ E − T δS) |T , and find that δ F |T may be expressed by the following types of work involved when h changes by the small amount δh: δ F |T = γT L δ A − γT A δ A + cghδV. Here δ A = 2π Rδh and δV = π R 2 δh, i.e. δ F |T = 2π R (γT L − γT A ) + π R 2 cgh δh.
(2.120)
(2.121)
2.3 Free Energy and Free Enthalpy
57
The quantities γT L and γT A are the surface tensions of the tube-liquid and tube-air interface, respectively, cf. (1.13). The last term in Eq. (2.120) is the negative of the work done by the system when raising the liquid volume δV to the height h. Notice that c is the mass density of the liquid, and g is the magnitude of the gravitational acceleration. In the above equations h is a parameter not controlled from outside the system. We are not doing work to the overall system nor do we extract work. The equilibrium height may be affected from the outside only by altering the temperature, changing the surface tension or the liquid density. But here T is held constant together with the overall volume of the system. Thus we may apply Eq. (2.118) to the above free energy, i.e. we find the equilibrium value of h by minimization of the above free energy with respect to h. This means we simply set the term in square brackets equal to zero: 0 = (γT L − γT A )2π R + cghπ R 2 .
(2.122)
Solving for h yields the equilibrium height h=
2(γT A − γT L ) . cg R
(2.123)
Because h depends on the sign of γT A − γT L , it may be positive or negative. A second solution to this problem follows via the force balance 2π R γ cos θ = γT A − γT L = m g, where m = cπ R 2 h (cf. Fig. 2.14). This implies that the above equation also can be written in terms of the liquid-air surface tension, γ , and the contact angle, θ , h=
2γ cos θ cg R
(2.124)
(e.g. γ = 0.0728 N/m for water-air at 20 ◦ C). A nice discussion of capillarity and wetting phenomena can be found in de Gennes et al (2004) (Pierre-Gilles de Gennes, Nobel Prize in chemistry for his contributions to the theory of polymers and liquid crystals, 1991.).
Example: Dielectric Liquid in a Plate Capacitor. The next, partially related problem is illustrated in Fig. 2.15. A plate capacitor is in contact with a liquid possessing again the mass density c and the dielectric constant εr . If a constant voltage, φ, is applied to the capacitor, the liquid rises to a certain equilibrium height between the capacitor plates. This is what we want to calculate.
58
2 Thermodynamic Functions
Fig. 2.14 Capillary rise TA
TL
h
Fig. 2.15 Dielectric liquid in a plate capacitor
h
However, first we must discuss the relation of the free energy, F, to the electric work expressed in Eq. (1.25) …to be continued… We start from δ F = δ(E − T S) = δ E − SδT − T δS
= δ E − SδT − δ E − = −SδT +
dV
E · δ D + ... dV 4π
E · δ D , 4π
i.e. δ F = −SδT +
dV
E · δ D . 4π
(2.125)
Momentarily we use only the electric field contribution from Eq. (1.21). The dots in the second line indicate that there are other types of work in general, which here either do not occur (e.g., chemical work) or can be neglected (e.g., volume change) or will be added later (work against the gravitational field). Also notice that
2.3 Free Energy and Free Enthalpy
59
E ∂ f , = T,V,... 4π ∂D
(2.126)
where f is a free energy density. However, this is not Thus we find that in the present case we have F = F(T, D). appropriate here—why? Notice that the scalar potential (voltage drop), φ is related to the (mean) electric field in the capacitor via . E = −∇φ
(2.127)
The equation applies to both the filled and the empty part of the capacitor. This means that if we hold the voltage on the capacitor plates constant, we also hold the E-field between the plates constant. If we want to minimize F = F(T, D; h) with respect to h, in order to compute the equilibrium height, we must do this keeping T and D fixed. But D is not the same as E, which is constant in the present setup. Thus we ˜ h) instead. need a new function F˜ = F(T, E; We try the Legendre transformation E · D (2.128) F˜ = F − d V 4π and obtain E · δ D δ E · D − dV 4π 4π δE · D , = −SδT − d V 4π
δ F˜ = δ F −
dV
i.e. δ F˜ = −SδT −
dV
δ E · D . 4π
(2.129)
˜ h). We now continue our example. Indeed we have F˜ = F(T, E; … The explicit E-dependent contribution to F˜ |T is obtained by integration ˜ E|T = −εr E V /(4π ) with respect to E from of the differential relation ∂ F/∂ zero to the actual field strength. Notice that we assume a constant field strength = εr E (This equation is used in sevbetween the capacitor plates as well as D eral places throughout this book, which means that the attendant calculations do depend on its validity!). Notice also that the field strength is the same with and without the dielectric. Analogous to Eq. (2.120) we collect all relevant work terms expressing δ F˜ |T via
60
2 Thermodynamic Functions
δ F˜ |T = − r
E 0
d E · E δV + 4π
E 0
d E · E δV + cghδV 4π
(2.130)
The first term is due to a volume increase, δV , of liquid between the capacitor plates. The second term is due to the corresponding reduction of vacuum (or air). Following the same reasoning as in the previous example, here the system includes the capacitor, the liquid, and the earth, we find the equilibrium height by setting the right side of the above equation equal to zero: 0=−
1 (εr − 1)E 2 + cgh. 8π
(2.131)
Solving for h yields h=
(εr − 1) E 2 . 8π cg
(2.132)
Remark: Somebody may comment that the introduction if F˜ is not necessary, be and E here are related via a constant εr . This is not true. Using F instead cause D ˜ of F changes the sign of the first term in brackets in Eq. (2.131) and there would be no sensible solution. Before leaving this subject we study the same problem from another angle. We detach the capacitor from its voltage supply. This means that we switch from φ = const to a situation where Q = const. Here ±Q is the charge on the capacitor plates. What happens to Eq. (2.132)? Figure 2.16 illustrates the situation. At (a) we have the boundary conditions ( E I − E I I ) × n = 0
and
I − D I I ) · n = 0, (D
(2.133)
where n is a unit vector perpendicular to the interface between liquid and vacuum (or air) at (a). At (b) we have instead E I × n = 0
and
I · n = 4π σ I . D
(2.134)
Now n is perpendicular to the capacitor plates. If we move (b) up into region II the boundary conditions become E I I × n = 0
and
I I · n = 4π σ I I . D
(2.135)
2.3 Free Energy and Free Enthalpy
61
Fig. 2.16 Different boundaries in the case of the liquid inside the plate capacitor
Q II
-Q
z
a
x I
h b
H
σ I and σ I I are the surface charge densities in these respective regions. From I = εr E I and D I I = E I I Eq. (2.133) we can see that E x,I = E x,I I . Because D we can conclude from Eqs. (2.134) and (2.135) that σ I = εr σ I I ,
(2.136)
i.e. the surface charge densities in the two regions are not equal. We can use this result to obtain Q = σI L y h +
σI L y (H − h) εr
or
σI =
εr Q/L y , (εr − 1)h + H
(2.137)
where L y is the width of the capacitor in y-direction. We now work out
E · D E · D − dV 8π f illed 8π empt y Vcapacitor Vcapacitor
A A 2 2 2 hεr E x,I H E x,I + (H − h)E x,I = I − I 8π 8π A (εr − 1)h E 2 . = 8π dV
(2.138)
In the last step we have use E x,I = E x,I I = E. This shows that we can treat this case analogous to the fixed potential case above. The result of course is the same as before. However, this time we must express E 2 via the total charge Q. Example: Euler Instability. The third example in this series is illustrated in Fig. 2.17. The figure shows a thin plate in a vice subject to a compressive force, f c . This systems encompasses the plate only. The pressure exerted by the vice is controlled from the outside. In this case there is an elastic Fel |T , i.e.
62
2 Thermodynamic Functions
Fel |T =
εh 3 24
d xd y plate
∂ 2ζ ∂x2
2 .
The quantity h is the plate’s thickness, ε is the elastic modulus of the plate’s material. The function ζ = ζ (x) describes the shape of the plate when looked at along the y-direction. The expression on the right side of the equation is the work done by the plate’s internal forces, when it is bent into the shape described by ζ (x). This result of continuum theory of elasticity may be found in Landau et al. (1986). What is the equilibrium shape of the plate depending on the applied compressive force? To find the answer we now employ relation Eq. (2.117), i.e. δ Fel |T +δw ≤ 0.
(2.139)
Here w is the negative of the work done by the vice, i.e. wc = −w. We compute wc by adding up the total displacement parallel to the direction of the force from infinitesimal increments (cf. the inset in Fig. 2.17):
ds − d x =
dx2
1 + (dζ (x)) − d x ≈ d x 1 + 2
2
∂ζ (x) ∂x
2
−1 .
Thus wc =
1 fc 2
dx
∂ζ ∂x
2 =
1 σ 2
dV V
∂ζ ∂x
2 .
V is the volume of the plate, and σ is the stress equal to f c /(h L y ), where L y is the extend of the plate in y-direction. We search for the minimum of the left side in Eq. (2.139) by “offering” suitable shape alternatives to the plate, i.e. we carry out a variation (or minimization) in terms of ζ :
0 = δζ (Fel − wc ) = δζ
dV
ε 2 z 2
∂ 2ζ ∂x2
2
σ − 2
∂ζ ∂x
2 .
2.3 Free Energy and Free Enthalpy
We write
1 h
h/2
−h/2 dzz
1 δζ 2
2
=
63
1 2 12 h
≡ I and thus
d xd y εI
∂ 2ζ ∂x2
2
−σ
∂ζ ∂x
2 = 0.
(2.140)
Carrying out the variation we obtain
d xd y εI
∂ 2ζ ∂x2
∂2 δζ ∂x2
−σ
∂ζ ∂x
∂ δζ = 0. ∂x
Partial integration using δζ (x) = 0 at ±L/2 finally yields the differential equation for the “shape function”: εI
∂ 4ζ ∂ 2ζ + σ = 0. ∂x4 ∂x2
Making the Ansatz ζ (x) = ζ0 sin (q x) we find 2 . εIq 4 − σ q 2 = 0 or σcrit = εIqmin
The planar solution is q = 0, whereas the bent plate corresponds to q > 0. The smallest possible non-vanishing q-value, qmin , defines the critical stress limit, σcrit , at which the transition from planar to bent occurs spontaneously. This value of qmin depends on the boundary conditions. Here we have qn = (π/L)n, where n = 1, 2, . . . and therefore σcrit =
π 2 εI . L2
(2.141)
In the literature this phenomenon is called Euler buckling. The problem may be modified by embedding the plate into an elastic medium. This results in larger values for qmin depending on the medium’s stiffness (Young’s modulus).
Remark: Notice that the above system has the freedom to decide to which side it buckles. This phenomenon is a spontaneous symmetry breaking.
64
2 Thermodynamic Functions
fc
x
L
z y x d
dx
ds
Fig. 2.17 Buckling of a thin plate
2.3.2 Maxwell Relations Equating the right sides of Eqs. (2.16) and (2.17) as well as the right sides of Eqs. (2.101) and (2.102) we have used that both F and G are state functions. The resulting formulas, Eqs. (2.18) and (2.103), are examples of so called Maxwell relations. It is easy to construct more Maxwell relations via the following recipe. Take any state function g and any pair of variables, x and y, it depends upon. If g = pd x +qdy then ∂q ∂ p = ∂y x ∂x y
(2.142)
yields a Maxwell relation. In general we may use the differential relations in Appendix A.2 to generate even more differential relations between thermodynamic quantities, i.e. more Maxwell relations. Example: Relating C V and C P . A nice exercise useful for practicing the “juggling” of partial derivatives, which is so typical for thermodynamics, is the derivation of the general relation between C V and C P . We start from C P − CV = Using G = H − T S and
∂ H ∂ E − . ∂T P ∂T V
(2.143)
2.3 Free Energy and Free Enthalpy
−S=
65
∂G ∂ H ∂ S = −S−T ∂T P ∂T P ∂T P
(2.144)
as well as F = E − T S and −S=
∂ E ∂ S ∂ F = −S−T ∂T V ∂T V ∂T V
(2.145)
CV = T
∂ S ∂T V
(2.146)
CP = T
∂ S . ∂T P
(2.147)
yields
and
With ∂ S (A.1) ∂ S ∂ S ∂ V = + ∂T P ∂T V ∂V T ∂T P
(2.148)
we have ∂ S ∂ V ∂V T ∂T P (2.18) ∂ P ∂ V = T ∂T V ∂T P ∂ P ∂ V (A.3) = −T ∂V T ∂T P α2 (2.5),(2.6) = TV P. κT
C P − CV = T
(2.149)
Using Eq. (2.9) we find for an ideal gas C P − C V = n R, which we had used in the calculation of the air temperature profile on p. 48.
(2.150)
66
2 Thermodynamic Functions
Example: Adiabatic Compressibility. Another exercise similar to the previous one is the derivation of κ S , the adiabatic compressibility. This quantity is defined in Eq. (2.89), i.e. κS = −
1 ∂ V . V ∂P S
We transform the right side via ∂ V (A.1) ∂ V ∂ T ∂ V = + . ∂P S ∂ T P ∂ P S ∂ P T (2.5)
(2.151)
(2.6)
= V αP
= −V κT
The remaining unknown derivative is ∂ T (A.3) ∂ T ∂ S ∂ T T = − = ∂P S ∂S P ∂P T ∂ S P C P (2.147)
=
∂ ∂G . T P T ∂ P ∂
= ∂ V =V α P ∂T
(2.152)
P
Putting everything together yields κ S = κT − T V
α 2P CP
(2.153)
or if we combine this equation with Eq. (2.149) κS CV = . κT CP
(2.154)
Again we can calculate κ S for an ideal gas. The result is κS =
z−1 1 . z P
(2.155)
The quantity z = C V /(n R) + 1 was introduced in the context of Eq. (2.77) (cf. its discussion in the footnote on p. 48). Example: Electric Field Effect on C V . What is the effect of an electric field on C V ? We find the answer by working from
2.3 Free Energy and Free Enthalpy
∂ ∂ E
∂ 2 F˜ ∂ T 2 ρ, E
67
T,ρ
=
∂2 ∂T 2
(2.146)
= −T −1 C V, E
∂ F˜ T,ρ ∂ E
ρ, E
,
(2.156)
(2.129)
= −V D/(4π )
i.e. ∂C V, E T V ∂2 D = , 4π ∂ T 2 ρ, E ∂ E T,ρ
(2.157)
= εr E where ρ = N /V and assuming constant fields throughout V . With D we may easily integrate this equation from zero to the final field strength, which yields C V (E) = C V (0) +
T V E 2 ∂ 2 εr . 8π ∂ T 2 ρ
(2.158)
Remark: A similar strategy helps to find corresponding expressions for κT or α P . Example: Electrostriction. Here we want to show the validity of E ∂εr 1 ∂ρ = ρ ∂ E μ,T 4π ∂ P E,T
(2.159)
(cf. Frank 1955). Imagine a plate capacitor completely submerged in a dielectric liquid. The dielectric constant of the liquid is εr . The size of the capacitor is small compared to the extend of the liquid reservoir. This means that far from the capacitor the latter has no effect on the chemical potential, μ, of the liquid, i.e. μ is constant. In addition the temperature, T , is held constant as between the capacitor plates will affect the well. The mean electric field, E, density, ρ = N /V , inside the capacitor. This is the expression on the left. The change does depend on electric field strength, E, and the derivative of εr with respect to pressure, P. Our starting point is relation (A.3), i.e.
68
2 Thermodynamic Functions
1 ∂ρ ∂μ 1 ∂ρ =− . ρ ∂ E μ,T ρ ∂μ E,T ∂ E ρ,T
(2.160)
(∗)
= ρκT
Note that (*) follows via ρ∂/∂ρ = −V ∂/∂ V and using ∂G/∂ V |T,N ,E = −1/κT derived below, cf. (2.183). We continue with ∂μ ∂ ∂ F˜ ∂ ∂ F˜ E ∂εr (2.129) = = = − .(2.161) ∂ E ρ,T ∂ E ∂ N T,V,E ρ,T ∂ N ∂ E ρ,T T,V,E 4π ∂ρ E,T The final ingredient is ρ
∂εr 1 ∂εr ∂εr (A.2) ∂ P = ρ = . ∂ρ E,T ∂ρ E,T ∂ P E,T κT ∂ P E,T
(2.162)
Combination of the last three equations yields the desired result. In order to estimate the magnitude of this effect we integrate Eq. (2.159) assuming that the derivative on the right side can be replaced by its value at zero field strength, i.e. E 2 ∂εr ρ ≈ . ρ 8π ∂ P o,T
(2.163)
√ The transition to SI-units is accomplished by replacing E with 4π εo E, i.e. in SI-units Eq. (2.163) becomes ρ/ρ ≈ (εo /2)E 2 ∂εr /∂ P|o,T . Because εo is a small constant (∼10−11 in these units), an appreciable effect requires rather high fields and/or conditions for which the derivative becomes large.13
2.4 Extensive and Intensive Quantities An important concept is the characterization of thermodynamic quantities as either intensive or extensive. Imagine two identical containers filled with the same kind and amount of of gas at the same temperature. What happens if we bring the two containers in contact and allow the gas to fill the combined containers as if it was one. Obviously temperature and pressure do not change. The quantities T and P therefore are called intensive. Volume on the other hand is doubled. Mathematically this means V ∝ n. Such quantities are said to be extensive. n itself is therefore extensive. Thus far: 13
In liquid water ∂ ln εr /∂ P|o,T ≈ 5 · 10−5 /bar .
2.4 Extensive and Intensive Quantities
69
• intensive: T , P, . . . • extensive: V , n, . . . The ratio of two extensive quantities, e.g. n/V , again is intensive of course. Another intensive quantity is the chemical potential, μ. Whether one mole of material is added to a large system or to twice as large a system should not matter.14 This however has implications for the free enthalpy, G. According to Eq. (2.107) we have for a one-component system dG T,P,... = μdn, (2.164) where . . . stands for other intensive variables in addition to T and P. Because μ is intensive and dn is extensive, we conclude that dG is extensive also. By adding (or integrating over) sufficiently many differential amounts of material, n = dn, we find the important relation G(T, P, n, . . . ) = μn.
(2.165)
For a K -component system this becomes G(T, P, n 1 , . . . , n K , . . . ) =
K
μi n i .
(2.166)
i=1
K
Equating dG on the left with d(
i=1 μi n i )
− SdT + V d P + · · · +
K
on the right, i.e.
μi dn i =
i=1
K
(μi dn i + n i dμi ),
(2.167)
i=1
yields −SdT + V d P + · · · −
K
n i dμi = 0,
(2.168)
i=1
the Gibbs-Duhem equation. The significance of this equation will become clear in many examples to come. Remark 1: Suppose we consider the potential energy U of a system consisting of N pairwise interacting molecules. Disregarding their spatial arrangement we may ∞ write U ∼ (N 2 /2)V −1 a drr 2−n . The factor N 2 ≈ N (N − 1)/2 is the number of distinct pairs, V is the volume, a is a certain minimum molecular separation, and r −n
14
Momentarily we talk about one-component systems and not about mixtures.
70
2 Thermodynamic Functions
is the leading distance dependence of the molecular interaction. If n > 3 the integral is finite and consequently U/V ∝ ρ 2 , where ρ is the number density molecules. This means that U (and also E) is extensive by our above definition. However, if n ≤ 3 the situation is more complex. Now it is necessary to include the spatial and orientational correlations between the molecules. In general these conspire to yield an extensive U. An exception is gravitation, where U/V ∼ ρ 2 V 2/3 including an additional shape dependence. Remark 2: Looking at the two Eqs. (2.164) and (2.165) we may wonder whether one can apply the same argument to d F T,V,... = μdn, valid according to Eq. (2.106). This immediately leads to F = G, which clearly is incorrect! The point is that we cannot keep the volume constant and simultaneously add up increments dn to the full n. Therefore this procedure does not work for d F T,V,... . Example: Partial Molar Volume. Consider a binary liquid mixture (A and B) at constant temperature and pressure. The volume change due to a differential change of the composition is ∂ V ∂ V dn A + dn B = v A dn A + v B dn B .(2.169) dV = ∂n A T,P,n B ∂n B T,P,n A The quantities v A and v B are called partial molar volumes. In this sense μi is the partial molar free enthalpy of component i. Now we argue as in the case of Eq. (2.164). V is extensive and so is n. Therefore we may ad up d V ’s to the full V at constant T and P and thus V = vAn A + vB n B .
(2.170)
However, the quantity of interest here is not V but the volume difference upon mixing, V . The volumes of the pure substances are vi∗ n i , where vi∗ = ∂ V /∂n i |T,P,n j =0(i= j) (i, j = 1, 2) and thus V = (v A − v ∗A )n A + (v B − v ∗B )n B .
(2.171)
Notice that v A and v B are not independent. To see this we simply must realize that we can carry out the steps from Eq. (2.166) to the Gibbs-Duhem equation (2.168) with G replaced by V and μi replaced by vi . At constant T and P this means dv A n A + dv B n B = 0.
(2.172)
2.4 Extensive and Intensive Quantities
71
In addition we notice that this may immediately be extended to more than two components. And this is not the end, because the same reasoning applies to every extensive quantity = (T, P, n 1 , n 2 , . . . ), i.e. K = (φi − φi∗ )n i ,
(2.173)
i
and K
n i dφi = 0
(T, P = constant),
(2.174)
i=1
where φi = ∂/∂n i |T,P,n j (i= j) are the respective partial molar quantities. Here stands for extensive thermodynamic quantities like V , H , C P , . . . .
2.4.1 Homogeneity Before leaving this subject we look at it briefly from another angle. In mathematics a function f (x1 , x2 , . . . , xn ) is said to be homogeneous of order m if the following condition is fulfilled: f (λx1 , λx2 , . . . , λxn ) = λm f (x1 , x2 , . . . , xn ).
(2.175)
Thus we may consider the extensive quantities free energy and free enthalpy as first-order homogeneous functions in V , n i and n i respectively, i.e. F(T, λV, λn 1 , λn 2 , . . . ) = λF(T, V, n 1 , n 2 , . . . ) G(T, P, λn 1 , λn 2 , . . . ) = λG(T, P, n 1 , n 2 , . . . ).
(2.176) (2.177)
Differentiating Eq. (2.176) on both sides with respect to λ yields ∂ F ∂ F dF = V+ n i = F. dλ ∂(λV ) T,n 1 ,n 2 ,... ∂(λn i ) T,V,n k(=i) K
(2.178)
i=1
For λ = 1 this becomes F=
K ∂ F ∂ F (2.109),(2.166) V+ ni = −P V + G (2.179) ∂ V T,n 1 ,n 2 ,... ∂n i T,V,n k(=i) i=1
72
2 Thermodynamic Functions
in agreement with Eq. (2.114). This also implies μi =
∂ F , ∂n i T,V,n k(=i)
(2.180)
i.e. the generalization of Eq. (2.110) to more than one component. Differentiating Eq. (2.177) on both sides with respect to λ and setting λ = 1 reproduces Eq. (2.166). Clearly, we may apply the same idea to other extensive thermodynamic functions like S(E, V, n 1 , n 2 , . . .), i.e. λS(E, V, n 1 , n 2 , . . .) = S(λE, λV, λn 1 , λn 2 , . . .), E(T, V, n 1 , n 2 , . . .), i.e. λE(T, V, n 1 , n 2 , . . .) = E(T, λV, λn 1 , λn 2 , . . .), or others. Likewise we may consider the intensive quantities as zero-order homogeneous functions in their extensive variables, e.g. P(T, V, n 1 , n 2 , . . .) = P(T, λV, λn 1 , λn 2 , . . .).
(2.181)
Differentiating with respect to λ on both sides and subsequently setting λ = 1 yields 0=
K ∂ P ∂ P V+ ni . ∂ V T,n 1 ,n 2 ,... ∂n i T,V,n k(=i)
(2.182)
i=1
Using P = −∂ F/∂ V |T,n 1 ,n 2 ,... and changing the order of differentiation we find 1 ∂G =− . ∂ V T,n 1 ,n 2 ,... κT
(2.183)
This easily is verified by insertion of Eq. (2.114) and subsequent differentiation. The concept of homogeneity does not produce otherwise unattainable relations, but it is an elegant means to compute them. We revisit homogeneity in a generalized form on p. 140 in the context of continuous phase transitions, where again it proves useful.
Chapter 3
Equilibrium and Stability
3.1 Equilibrium and Stability via Maximum Entropy 3.1.1 Equilibrium The first row of boxes shown in Fig. 3.1 depicts a number of identical systems differing only in their internal energies, E ν , volumes, Vν , and mass contents, n ν . The boundaries of the systems allow the exchange of these quantities between the systems upon contact. The second row of boxes in Fig. 3.1 illustrates this situation. All (sub-)systems combined form an isolated system. We ask the following question: What can be said about the quantities xν , where x represents E, V or n, after we bring the boxes into contact and allow the exchanges to occur? According to our experience the exchange is an irreversible spontaneous process and therefore relation Eq. (1.50) applies to the entropy of the overall system. We can expand the entropy of the combined systems, S, in a Taylor series in the variables E ν , Vν , and n ν with respect to its maximum, i.e. ∂ Sν o ∂ Sν o ∂ Sν o E ν (3.1) S = So + + Vν + n ν ∂ E ν Vν ,n ν ∂ Vν E ν ,n ν ∂n ν E ν ,Vν ν ∂ o ∂ o ∂ o 1 E ν + + Vν + n ν 2 ∂ E ν Vν ,n ν ∂ Vν E ν ,n ν ∂n ν E ν ,Vν ν ,ν ∂ Sν o ∂ Sν o ∂ Sν o , × E ν + Vν + n ν ∂ E ν Vν ,n ν ∂ Vν E ν ,n ν ∂n ν E ν ,Vν o where S = μ Sμ (E μ , Vμ , n μ ). The quantity S is the maximum value of the entropy. Notice that this quantity is somewhat hypothetical. The usefulness of this approach relies on the differences between time scales on which certain processes take place.
R. Hentschke, Thermodynamics, Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-36711-3_3, © Springer-Verlag Berlin Heidelberg 2014
73
74
3 Equilibrium and Stability
Leaving a cup of hot coffee on the table, we expect to find coffee at room temperature upon our return several hours later. If we come back after some weeks of vacation the coffee has vanished, i.e. the water has evaporated. Only the dried remnants of the coffee remain inside the cup. After waiting for a much longer time, how long depends on numerous things including the material of the coffee cup, the cup itself has crumbled into dust. However, if we are interested merely in the initial cooling of the coffee to room temperature, we may neglect evaporation and we may certainly neglect the deterioration of the cup itself. In this sense we shall use the expression of equilibrium. For all practical purposes equilibrium is understood in a “local” sense, i.e. the time scale underlying the process of interest is much shorter than the time scale underlying other processes influencing the former. In the case at hand equilibrium means that all variables, E ν , Vν , and n ν , have assumed the values E νo , Vνo , and n oν corresponding to maximum entropy. However, we may impose deviations from these values in each subsystem, xν , as illustrated in the bottom part of Fig. 3.1. The long dashed line indicates the equilibrium value(s), which is the same in all (identical) systems. The short dashed lines indicate the imposed deviations from equilibrium in each system, xν . Because the whole system is isolated, we have the condition(s)
xν = 0.
(3.2)
ν
Equation (3.1) is nothing but a Taylor expansion of S to second order in the xν , which we can freely and independently adjust except for the condition(s) Eq. (3.2). Here the value of S o is of no interest to us. But already the linear terms, i.e the first sum, leads to important conclusions. If for the moment we consider two subsystems only, i.e. ν = 1, 2, then the condition of maximum entropy yields
E1, V1, n1
E2, V2, n2
E3, V3, n3
E ,V ,n
x x1
x2
x3
x
Fig. 3.1 Identical systems initially differing only in their internal energies, volumes, and mass content
3.1 Equilibrium and Stability via Maximum Entropy
75
∂ S1 o ∂ S2 o − ∂ E 1 V1 ,n 1 ∂ E 2 V2 ,n 2 ∂ S1 o ∂ S2 o + V1 − ∂ V1 E 1 ,n 1 ∂ V2 E 2 ,n 2 ∂ S1 o ∂ S2 o , + n 1 − ∂n 1 E 1 ,V1 ∂n 2 E 2 ,V2
0 = E 1
where we have used S = and (1.55) this becomes E 1
1 1 − T1 T2
ν
(3.3)
Sν (E ν , Vν , n ν ) and Eq. (3.2). Via the Eqs. (1.52), (1.53),
+ V1
P1 P2 − T1 T2
− n 1
μ1 μ2 − T1 T2
= 0.
Because E 1 , V1 , and n 1 are arbitrary, we conclude that T = T1 = T2 P = P1 = P2
(3.4) (3.5)
μ = μ1 = μ2
(3.6)
at equilibrium. These conditions of course may be generalized to an arbitrary number of subsystems. The latter in general are different regions in space within a large system. In some cases different regions in space may contain distinct phases. An example is ice in one region of space and liquid water in an adjacent region. One and the same material may occur in different phases depending on thermodynamic conditions. A phase is a homogeneous state of matter. Each phase usually differs from another phase by certain clearly distinguishable bulk properties. Ice, for instance, has a lower symmetry than liquid water. At coexistence, defined by the above conditions, ice has a lower density than liquid water etc. Changing from one phase to another often, but not always, is accompanied by a discontinuous change of certain thermodynamic quantities. We shall discuss phase transformations in detail below. Equation (3.6) is derived for a one-component system. Of course we can extend our reasoning to a K -component system, which yields (1)
μi
(2)
= μi
(3.7)
(i = 1, . . . , K ). Here (1) and (2) are the subsystem indices. Again (1) and (2) may refer to different phases, i.e. at equilibrium the chemical potential of each component is continuous across the phase boundary. In particular if there are phases, each considered to be a subsystem, we find (ν)
μi
(μ)
= μi ,
(3.8)
76
3 Equilibrium and Stability
Fig. 3.2 Hypothetically coexisting phases
Gas
Gas
Liquid
Liquid
Solid
Solid
Flubber
Flubber
where ν, μ = 1, 2, . . . , .
3.1.2 Gibbs Phase Rule In general there are K components in different phases (solid, liquid, gas, . . .) at constant T and P. We may ask: What is the maximum number of coexisting phases at equilibrium? Or to be pictorial, is the situation in Fig. 3.2 possible, where a onecomponent system contains four coexisting phases—“Gas”, “Liquid”, “Solid”, and “Flubber”? We assume a system containing K components and coexisting phases. Each phase may be considered a subsystem in the above sense. The state of each phase ν is then determined by its temperature T (ν) , its pressure P (ν) , and its composition (ν) (ν) {n (ν) 1 , n 2 , . . . , n K }. All in all we must specify (K + 2)
(3.9)
quantities. On the other hand, equilibrium, as we have just discussed, imposes certain constraints. In the case of two subsystems (now phases) and just one component we had to fulfill the Eqs. (3.4) to (3.6). In the case of phases and K components we have T (1) = T (2) = . . . = T () P (1) = P (2) = . . . = P () (1) (2) () μ1 = μ1 = . . . = μ1 (1) (2) () μ2 = μ2 = . . . = μ2 .. . (1)
(2)
()
μK = μK = . . . = μK and therefore
3.1 Equilibrium and Stability via Maximum Entropy
77
( − 1)(K + 2)
(3.10)
constraints. In addition there are constraints having to do with the total amount of material in each phase. In the one-component system illustrated in Fig. 3.2 we may for instance insert a diagonal partition without physical effect—it is a key assumption that the shape of our container has no influence on the type of phases present. This means that the total amount of material in a phase ν, n
(ν)
=
K
(ν)
ni ,
i=1
does not affect the phase coexistence. This yields
(3.11)
constraints. The net number of adjustable quantities, i.e. the number of overall adjustable quantities, Eq. (3.9), minus the number of constraints, Eqs. (3.10) and (3.11), is called the number of degrees of freedom, Z. Thus Z = K − + 2 ≥ 0.
(3.12)
Applied to our above system we find 1 − 4 + 2 < 0. This means that four phases cannot coexist simultaneously in a one-component system. The maximum number of coexisting phases in a one-component system is three—but thermodynamics does not specify which three phases. Relation Eq. (3.12) is Gibbs phase rule. Remark 1: Our reasoning is based on subsystems whose state is determined by temperature, pressure, and composition. However, we may also include external electromagnetic fields requiring recalculation of the degrees of freedom. Remark 2: Below we shall discuss in more detail what we mean by component. This in turn will affect the statement of the phase rule (cf. p. 99).
3.1.3 Stability We now return to Eq. (3.1) and focus on the second order term. According to our discussion the linear term vanishes at equilibrium. In addition, in the second order term, the cross contributions (ν = ν ) also vanish and therefore S = S − S o is given by
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3 Equilibrium and Stability
∂ o ∂ o ∂ o 1 S = + Vν + n ν E ν 2 ν ∂ E ν Vν ,n ν ∂ Vν E ν ,n ν ∂n ν E ν ,Vν ∂ Sν o ∂ Sν o ∂ Sν o +V +n × E ν ν ν ∂ E ν V ,n ∂ Vν E ,n ∂n ν E ,V
ν ν
ν ν
ν ν =
1 2
=1/T
=P/T
=−μ/T
[. . .]Sν
ν
1 1 1 − Sν [. . .]T + [. . .](T Sν ) 2 ν T T 1 = (−Sν Tν + Pν Vν − μν n ν ) , 2T ν
=
(3.13)
where we have used T S = E + PV − μn,
(3.14)
cf. (1.51). Equation (3.13) quite generally expresses the entropy fluctuations via the corresponding fluctuations in the subsystems. Now we choose the variables T , V , and n, which yields ∂ Sν o ∂ Sν o ∂ Sν o Tν + Vν + n ν Tν (. . .) = − ∂ Tν Vν ,n ν ∂ Vν Tν ,n ν ∂n ν Tν ,Vν ∂ Pν o ∂ Pν o ∂ Pν o + Tν + Vν + n ν Vν ∂ Tν Vν ,n ν ∂ Vν Tν ,n ν ∂n ν Tν ,Vν ∂μν o ∂μν o ∂μν o − Tν + Vν + n ν n ν ∂ Tν Vν ,n ν ∂ Vν Tν ,n ν ∂n ν Tν ,Vν CV ∂ Pν o ∂μν o Tν − =+ − Vν + n ν Tν T ∂ Tν Vν ,n ν ∂ Tν Vν ,n ν ∂ Pν o 1 ∂μν o + Tν − o o Vν − n ν Vν ∂ Tν Vν ,n ν V κT ∂ Vν Tν ,n ν ∂μν o ∂μν o ∂μν o + − Tν − Vν − n ν n ν ∂ Tν Vν ,n ν ∂ Vν Tν ,n ν ∂n ν Tν ,Vν
CV 1 ∂μν o ∂μν o 2 2 2 Tν − =− Vν − n ν − 2 n ν Vν . T V κT ∂n ν Tν ,Vν ∂ Vν Tν ,n ν We need to transform this equation one last time using
3.1 Equilibrium and Stability via Maximum Entropy
79
∂ Vν o ∂ Vν o ∂ Vν o T + P + n ν ν ν ∂ Tν Pν ,n ν ∂ Pν Tν ,n ν ∂n ν Tν ,Pν ∂ Vν o ≡ Vn,ν + n ν . ∂n
Vν =
ν Tν ,Pν
The quantity Vn,ν is the volume fluctuation at constant mass content. We obtain (. . .) = −
CV 1 ∂μν o 2 Vn,ν − n 2ν . Tν2 − T V κT ∂n ν Tν ,Pν
(3.15)
According to the second law S must be negative, because otherwise the fluctuations would grow spontaneously in order to increase the entropy. Therefore we find CV ≥ 0
κT ≥ 0
∂μ o ≥0 ∂n T,P
(3.16)
for the isochoric heat capacity, C V , the isothermal compressibility, κT , and the quano ∂μ tity ∂n . These relations are sometimes denoted as thermal stability, mechanical T,P
stability, and chemical stability.1 Consequently we also have ∂ 2 G 1 = − CP ≤ 0 2 ∂ T P,n T 2 ∂ F 1 = − CV ≤ 0 ∂ T 2 V,n T
∂ 2 G = −V κT ≤ 0 ∂ P 2 T,n ∂ 2 F 1 = ≥0 ∂ V 2 T,n V κT
(3.17) (3.18)
(0 ≤ C V ≤ C P !). Remark: The above condition for chemical stability can be generalized to K components by replacing the one-component terms, n ν . . ., in the derivation with their multicomponent versions, k n ν,k . . .. The result is K ∂μ j o n j n k ≥ 0. ∂n k T,P
(3.19)
j,k=1
The simplest way to get rid of the subsystem indices ν is to consider two subsystems only, i.e. n 1,k = −n 2,k = n k .
1 The conditions Eq. (3.16) are mathematical statements of Le Châtelier’s principle, i.e. driving a system away from its stable equilibrium causes internal processes tending to restore the equilibrium state.
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3 Equilibrium and Stability
3.2 Chemical Potential and Chemical Equilibrium 3.2.1 Chemical Potential of Ideal Gases and Ideal Gas Mixtures Pure gas: Based on the Gibbs-Duhem equation (2.168) we may write for the chemical potential of a one-component gas A at fixed temperature T (g)
μA
1 (g) T, PA∗ − μ A T, PA◦ = n
PA∗ PA◦
V d P.
(3.20)
The various indices do have the following meaning. The index (g) reminds us that we talk about gases. The index ∗ indicates that the gas is a pure or one-component gas and not one component in a mixture of gases. The other index ◦ indicates a reference pressure. In thermodynamics the chemical potential is not an absolute quantity. We rather compute differences between chemical potentials—often there is some standard state, defined by specifying temperature and pressure values, with respect to which the difference is calculated. If our gas is ideal the above equation becomes (g)
μA
(g) T, PA∗ − μ A T, PA◦ = RT
PA∗ PA◦
dP , P
i.e. (g)
μA
P∗ (g) T, PA∗ = μ A T, PA◦ + RT ln A◦ . PA
(3.21)
Mixture: For a mixture of K components at fixed T the Gibbs-Duhem equation yields K
n i dμi = V d P.
(3.22)
i=1
Here we can make use of Dalton’s law stating that for a mixture of ideal gases P=
K i=1
(g)
Pi and Pi = RT
ni . V
(3.23)
The quantities P and V are the total pressure and the total volume. The quantities Pi are called partial pressures. Pi is the contribution to the total pressure due to the (g) presence of n i moles of gas i. Thus Eq. (3.22) becomes
3.2 Chemical Potential and Chemical Equilibrium K
(g)
ni
(g)
dμi
81 (g)
− RT d ln n i
= 0.
(3.24)
i=1 (g)
In general this will be true only if dμi integration yields
(g)
(g)
μ A (T, PA ) − μ A
(g)
− RT d ln n i
= 0 for all i, which after
(g) nA T, PA∗ = RT ln (g) . n
(3.25)
The chemical potential difference on the left is between a mixture of a certain (g) K (g) , where n A = PA V /(RT ), and a pure A-gas at given composition {n i }i=1 K (g) PA∗ = P = RT n (g) /V . Because n (g) = i=1 n i , we may write (g)
(g)
(g)
(g)
μ A (T, PA ) = μ A
PA T, PA∗ + RT ln ∗ PA
(3.26)
(g) T, PA∗ + RT ln x A ,
(3.27)
or μ A (T, PA ) = μ A where we have used the definition (g) xA
(g)
nA = (g) . n
(3.28)
(g)
The quantity x A is the mole fraction of component A and K
(g)
xi
= 1.
(3.29)
i=1
Even though we use PA as the pressure argument in the chemical potentials on the left sides of Eqs. (3.26) and (3.27), the total pressure still is P, i.e. the total pressure is the same on both sides of these equations. Equations (3.26) and (3.27) describe the difference between the molar chemical potential of the A-component in an ideal mixture and the molar chemical potential of A in the pure and ideal A-gas at temperature T and identical overall pressure P. Combining Eq. (3.26) with Eq. (2.166) we may write down the free enthalpy of mixing for a K -component ideal gas, i.e. (g)
(g)
m G(T, P, n 1 , . . . , n K , . . .) = n (g) RT
K i=1
(g)
xi
(g)
ln xi .
(3.30)
82
3 Equilibrium and Stability
μ A(g)=μ
P* A
A(g)
(g)+RT ln (P*A/PoA)
equal μ
A(l)
(l)
Fig. 3.3 A pure gas A coexisting with its liquid
3.2.2 Chemical Potential in Liquids and Solutions Pure liquid: Thus far we have dealt with gases. The new situation is illustrated in Fig. 3.3. A pure gas A coexisting with its liquid. This may be achieved by partly filling a container with the liquid of interest. After closing the container tightly an equilibrium between liquid and gaseous A develops according to the conditions Eqs. (3.5) and (3.6).2 In particular according to Eq. (3.6) the chemical potentials of A must be the same in the gas and in the liquid, i.e. (l)
(g)
μ A (T, PA∗ ) = μ A (T, PA∗ ).
(3.31)
Here the index (l) indicates the liquid state and PA∗ now denotes the equilibrium gas pressure at coexistence. Assuming that the gas above the liquid is ideal we may obtain the right side of Eq. (3.31) by integrating from a reference pressure PA◦ just as in the case of Eq. (3.26), i.e. (l)
(g)
(g)
μ A (T, PA∗ ) = μ A (T, PA∗ ) = μ A
P∗ T, PA◦ + RT ln A◦ . PA
(3.32)
This result applies along the coexistence curve separating gas and liquid in the T P-plane. There should be such a curve according to the phase rule Eq. (3.12) applied to a one-component system. In this case Z = 1 − 2 + 2 = 1. We may vary one degree of freedom, T or P, which then fixes P or T , respectively. Figure 3.4 shows a partial sketch of the coexistence curve for a one-component system.3 Equation (3.32)
2 Thermodynamics does not predict the states of matter or describe their structure. Their existence, here gas and liquid, is an experimental fact, which we use at this point. 3 We shall show how to calculate this curve on the basis of a microscopic interaction model—the van der Waals theory.
3.2 Chemical Potential and Chemical Equilibrium Fig. 3.4 Partial gas-liquid coexistence curve in a onecomponent system
83
liquid
PA
b
PA
a gas
T
applies to point (a) but not to point (b) inside the liquid region. However, we can calculate the chemical potential at point (b) in the liquid via
1 (b) (l) (a) μ(l) A (T, PA ) = μ A (T, PA ) + (l) n
≡PA∗
(b)
PA
(a)
V (P)d P.
(3.33)
PA
We do not know V (P) in the liquid. But we do know from experience that the volume of a liquid changes little, compared to a gas, when the pressure is increased. Thus (a) we simply Taylor-expand V (P) around V (PA ), i.e. (a) (a) (a) V (P) ≈ V (PA ) 1 − κT (PA )(P − PA ) ,
(3.34)
where κT is the isothermal compressibility defined via Eq. (2.6). Typical liquid compressibilities are in the (G Pa)−1 -range. This means that the second term usually can be neglected, i.e. 1 (l) (b) (l) (a) (a) (b) (a) μ A (T, PA ) ≈ μ A (T, PA ) + (l) V (PA )(PA − PA ). n
≡PA∗
(3.35)
84
3 Equilibrium and Stability
Example: Relative Humidity. An experiment is carried out at temperature T pressure P and 50 % relative humidity—what does this mean? The relative humidity, ϕ, is defined via PD (T ) = ϕ Psat (T ).
(3.36) (a)
Psat (T ) is the saturation pressure of water at T , which is the pressure PA in Fig. 3.4. PD (T ), on the other hand, is the partial water vapor pressure in air at T and relative humidity ϕ · 100 % . Before we come to the actual problem, we want to get a feeling for relative humidity, i.e. we ask: what is the mass of water contained in cubic meter of air if the relative humidity is 40 %? We look up the vapor pressure from a suitable table, e.g. HCP. At T = 0 ◦ C and T = 20 ◦ C we find Psat = 0.006 bar and Psat = 0.023 bar, respectively. Using the ideal gas law, Psat V = n RT , we obtain the corresponding masses of water vapor, i.e. 4.8 and 17.3 g/m3 , on the coexistence line. The water content at ϕ = 0.4 is therefore 1.9 and 6.9 g/m3 . This is quite small compared to an approximate mass density of air of 1000 g/m3 . Figure 3.5 shows the gas-liquid coexistence or saturation line for water (solid line). The data are from HCP. On this line the relative humidity is 100 %. The dashed lines correspond to lines of constant humidity as indicated. The horizontal arrow indicates the cooling of air originally at 25 % relative humidity at constant pressure until the saturation line is reached. The temperature at which this happens and the water vapor starts to condense is called dew point. The vertical arrow indicates a portion of a drying process. Dry air increases its moisture contents. Subsequently it may be cooled and upon reaching the saturation line the vapor in the air condenses. By moving along the saturation line towards lower partial water pressure more water is removed form the air. Eventually heating of the air restores it to its starting point at low relative humidity. However, our real problem is the following. We are interested in the differ(g) ence between the chemical potential of water in the gas phase, μ H2 O (T, P), at a relative humidity ϕ and the chemical potential in the liquid phase of pure ◦ water, μ(l) H2 O (T, P), under the same conditions (for example T = 40 C and P = 1 bar). Such a question may arise when the water uptake in a material is measured by one experiment at fixed T , P, and ϕ in the gas phase or by another experiment via submerging the same material in liquid water at otherwise identical conditions. According to Eqs. (3.26) and (3.35) we have (g)
(g)
μ H2 O (T, PD ) ≈ μ H2 O (T, Psat ) + RT ln
PD Psat
3.2 Chemical Potential and Chemical Equilibrium
85
and (l) μ(l) H2 O (T, P) ≈ μ H2 O (T, Psat ) +
1 V (P)(P − Psat ). n (l)
(g)
With μ H2 O (T, Psat ) = μ(l) H2 O (T, Psat ) and using Eq. (3.36) we obtain for the (g)
(l)
difference μ H2 O (T ) ≡ μ H2 O (T, P) − μ H2 O (T, PD ) μ H2 O (T ) ≈ −RT ln ϕ.
(3.37)
Notice that the neglected term, i.e. n1(l) V (Psat )(P − Psat ), is small. With a liquid water molar volume of 18 cm3 and Psat (40 ◦ C) = 0.0737 bar we obtain ≈ 1.7 × 10−3 kJmol−1 . Because ϕ = 0.5, i.e. 50 % relative humidity, we find finally μ H2 O ≈ 1.8 kJmol−1 .
Remark 1: In the preceding discussion we have implicitly assumed a one-component system, i.e. neat water. However, relative humidity belongs to our everyday life. This means that we deal with air, a mixture which includes gaseous water as one particular component. In addition ordinary liquid water also contains a certain amount of each of the gaseous components which can be found in the air. Therefore we must ask, how much is the saturation line of neat water shown in Fig. 3.5 affected by the presence of other components. We shall answer this question on page 89 after we have discussed solutions. 0.30
PH2O [ bar ]
75 %
100 %
0.25 0.20
50 %
0.15 0.10
25 % 0.05 260
280
300
320
340
T [K] Fig. 3.5 Water saturation line including lines of constant humidity
360
86
3 Equilibrium and Stability 0.025 0.020
P [bar]
1100 m 0.015
70 %
0.010
50 %
2100 m 0.005
0
5
10
15
20
T [°C] Fig. 3.6 Cloud base for different humidities
Remark 2: Cloud Base. We want to estimate the lowest altitude of the visible portion of a cloud, i.e. the cloud base. The idea is as follows. Our study of the temperature profile of the troposphere (see p. 46) has resulted in the Eqs. (2.82) and (2.85) allowing to relate pressure, temperature, and corresponding height above see level for air. If we apply these two formulas to the partial pressure of water vapor in air at a given humidity, we can estimate at what temperature (if at all) the partial pressure in air will become equal to the saturation pressure of water. The resulting temperature may then be used to compute the height at which this happens. This is the height when the water vapor condenses an thus defines the cloud base. Neglecting the effect of the different molar weights of water and (dry air), we compute the partial water H 2O (T ). T = 20 ◦ C is the ground pressure via P = Po (T /To )3.5 , where Po = ϕ Psat o o temperature. The two dashed curves in Fig. 3.6 are for ϕ = 0.5 and ϕ = 0.7, i.e. 50 and 70 % relative humidity, respectively. The solid line in Fig. 3.6 is the saturation line for water. The two temperatures at which the curves intersect are converted into heights, i.e. ≈2100 and ≈1100 m. We notice that the cloud base is lower when the humidity is greater. Taking into account the molar weight difference mentioned above decreases these values by roughly 10 %. A sensitive quantity is To , i.e. decreasing To also decreases the cloud base. Solution: A more complex system is shown in Fig. 3.7. A binary solution of the components A and B is in equilibrium with its gas. The coexistence conditions are (l)
(g)
(l)
(g)
μ A (T, PA , PB ) = μ A (T, PA , PB )
(3.38)
and μ B (T, PA , PB ) = μ B (T, PA , PB ).
(3.39)
Assuming that the gas phase is an ideal mixture we make use of Eq. (3.26) to express this as
3.2 Chemical Potential and Chemical Equilibrium
A(g) + B(g)
87
μA(g)=μ
PA+PB
+RT ln (PA/Po)
equal A(l) + B(l)
μ (l)
Fig. 3.7 A binary solution of the components A and B in equilibrium with its gas (g)
μ(l) A (T, PA ) = μ A
PA T, PA∗ + RT ln ∗ PA
(3.40)
PB T, PB∗ + RT ln ∗ . PB
(3.41)
and (l)
(g)
μ B (T, PB ) = μ B
Note that ∗ refers to the same system with all B-moles in Eq. (3.38) replaced by A-moles and vice versa in Eq. (3.39). This means that Pi ∗ is the vapor pressure in the pure i system at coexistence. Therefore we may write PA (l) (l) μ A (T, PA ) = μ A T, PA∗ + RT ln ∗ PA
(3.42)
PB (l) ∗ μ(l) B (T, PB ) = μ B T, PB + RT ln ∗ . PB
(3.43)
and
Using partial pressures is somewhat inconvenient. There are two limiting laws, which are very useful here. The first is Raoult’s law, (l)
PA = PA∗ x A , (l))
(3.44)
(l)
valid if x A x B , i.e. the liquid is a (very) dilute solution of solute B in solvent (l)) (l) A.4 Notice that x A + x B = 1. Inserting Raoult’s law in Eq. (3.42) yields (l) (l) (l) μ A (T, PA ) = μ A T, PA∗ + RT ln x A .
4
The meaning of A and B can of course be interchanged.
(3.45)
88
3 Equilibrium and Stability
Raoult’s law is an example for a colligative property of the solution. Colligative properties of solutions depend on the number of molecules in a given amount of solvent and not on the particular identity of the solute. The second useful law is Henry’s law, (l)
PB = K B x B ,
(3.46)
where K B is called Henry’s constant. Henry’s constant is not a universal constant. It depends on the system of interest and on temperature. Henry’s law is valid in the same limit as Raoult’s law, i.e. x B 1. Inserting Eq. (3.46) into Eq. (3.43) yields KB (l) (l) (l) μ B (T, PB ) = μ B T, PB∗ + RT ln ∗ + RT ln x B . PB
(3.47)
The first two terms may be absorbed into the definition of a new hypothetical reference state with the chemical potential μ¯ B (l)
(l)
μ B (T, PB ) = μ¯ B (T ) + RT ln x B .
(3.48)
(l)
We call this reference state hypothetical, because μ B (T, PB ) = μ¯ B (T ) requires x B = 1, a concentration at which Henry’s law does not apply. Table 3.1 compiles Henry’s constant for oxygen, nitrogen, and carbon dioxide in water (based on solubility data in HCP). In the third column c H2 O is the mass density of the respective component in water in equilibrium with air at a pressure of 1 atm. The last column shows the corresponding mass density in air. Figure 3.8 shows the partial pressures in the two-component vapor-liquid system acetone-chloroform at T = 308.15 K (Ozog and Morrison 1983). Solid lines are polynomial fits to the data points. The long dashed lines illustrate Raoult’s law applied to the two components while the short dashed lines illustrate Henry’s law.
Table 3.1 Henry’s law applied to three gases in water Component
T [K ]
K H [109 Pa]
c H2 O [g/m 3 ]
cair [g/m 3 ]
O2
288.15 298.15 308.15 288.15 298.15 308.15 288.15 298.15 308.15
3.7 4.4 5.1 7.3 8.6 9.7 0.12 0.16 0.21
10 8.6 7.4 17 14 13 78 59 45
284 275 266 924 893 864 71 68 66
N2
C O2
3.2 Chemical Potential and Chemical Equilibrium
89
350 300
P [Torr]
250 200 150 100 50 0
0.0
0.2
0.4
0.6
0.8
1.0
xCHCl3 Fig. 3.8 Partial pressures in the two-component vapor-liquid system acetone-chloroform at T = 308.15 K
Remark: We now return to the question on p. 85—how much is the saturation line of neat water shown in Fig. 3.5 affected by the presence of other components? When water vapor coexists with liquid water we have (g)
(l)
μ H2 O (T, Psat , P) = μ H2 O (T, Psat , P) (3.40)
(g)∗
= μ H2 O (T, P) + RT ln
(3.49) Psat . P
Here P is the overall gas pressure and Psat is the partial water pressure at coexistence. The star indicates pure water. Let us assume we change P by an amount d P. This will alter the two sides of the above equation but the changes still will be the same on both sides: still ∗ ) ∂μ H2 O (T, Psat , P) ∂μ H2 O (T, P = Psat Psat . dP = d P + RT d ln T T ∂ P ∂ P P
(g)
(l)
(l) (T,Psat ,P) 2O
(g)∗ (T,P)=RT /P 2O
=v H
=v H
(3.50) (l)
Here v H2 O (T, Psat , P) is the partial molar volume of water in contact with air at (g)∗
pressure P, and v H2 O (T, P) is the same quantity for pure water vapor at pressure P. Under standard conditions (1 bar) we do not make a big mistake if we replace (l) (l)∗ v H2 O (T, Psat , P) by the same quantity for pure water, v H2 O (T, P). Thus we obtain (l)∗
v H2 O (T, P)d P ≈ RT d ln Psat .
(3.51)
∗ , the saturation pressure of pure water at T to Now we integrate the left side from Psat the ambient pressure of air (including water vapor), P. The corresponding integration
90
3 Equilibrium and Stability
∗ and P . This yields limits on the right side are also Psat sat ∗ v (l)∗ H2 O (T, P)(P − Psat ) ≈ RT d ln
Psat ∗ Psat
(3.52)
or (l)∗ v H2 O (T, P) Psat ∗ (P − Psat ) . ∗ ≈ exp Psat RT
(3.53)
Let us assume P = 1 bar = 105 Pa and T = 293 K. The saturation pressure of pure ∗ = 2338.8 Pa. In addition v (l)∗ ≈ 18 × 10−6 m 3 . water at this temperature is Psat H2 O And thus we find Psat −4 . ≈ exp 7 × 10 ∗ Psat
(3.54)
This means that the saturation pressure of water under ambient conditions is scarcely different from the saturation pressure of pure water at the same temperature.
3.3 Applications Involving Chemical Equilibrium Osmotic Pressure Figure 3.9 shows a beaker containing the pure liquid A. Immersed in the liquid is a tube with its lower end closed to the liquid by a membrane. The membrane allows A to permeate into the tube and vice versa. Inside the tube there is a binary mixture of two components A and B. The latter however is held back inside the tube by the membrane. What happens? As far as A is concerned the two subsystems, the pure solvent outside the tube and the binary mixture inside the tube, do exchange A-moles and therefore chemical equilibrium requires μ∗A (T, P) = μ A (T, P + , x A ).
(3.55)
The left side is the chemical potential of pure A outside the tube. The right side is the chemical potential of A inside the tube. Because we do not consider the gas phase we may omit the index (l) . The temperature is the same in both subsystems. The outside pressure is P, whereas the inside pressure is different i.e. P + . Why? Initially the tube may contain B only. Chemical equilibrium therefore requires flow of A across the membrane into the tube. For simplicity we assume that the
3.3 Applications Involving Chemical Equilibrium
91
Fig. 3.9 Sketch of a simple osmotic pressure experiment
μA(P+ )
h
membrane
μ*A(P)
initial surface level inside and outside the tube is the same and the density of A and B is the same as well. The pressure difference across the membrane, , can then be determined by measuring h (at equilibrium) and computing the force of gravitation exerted by the mass of material above the surface level of the surrounding A solvent. The reason for the sustained pressure difference is the membrane, which does not allow the chemical equilibration of B on both sides. Making use of the Gibbs-Duhem equation (2.168) we have μ∗A (T, P + ) ≈ μ∗A (T, P) +
1 ∗ V (P)(P + − P). n ∗A
(3.56)
Again it is assumed that incompressibility of the liquid is a good approximation, cf. (3.34). Combination of Eqs. (3.55) and (3.56) then yields μ∗A (T, P + ) −
1 ∗ V (P) ≈ μ A (T, P + , x A ), n ∗A
(3.57)
and according to Eq. (3.45), if x A x B , we obtain μ∗A (T, P + ) −
1 ∗ V (P) ≈ μ∗A (T, P + ) + RT ln x A . n ∗A
(3.58)
At first glance this may seem strange, because in Eq. (3.45) the pressure arguments of the chemical potentials are PA and PA∗ , which are presumably different. In general the chemical potential of a particular species does depend on temperature, total pressure, and composition. In Eq. (3.45) the total pressure is the same on both sides of the equation and does not show up explicitly in the lists of arguments. The composition dependence is expressed in terms of different partial pressures via (originally)Dalton’s law. Here the total pressure is included in the argument of the chemical potential and the composition is expressed in mole fractions. Having explained this we can write down the final result for given by ≈
RT V A,mol
xB,
(3.59)
92
3 Equilibrium and Stability
where we have used the molar volume of A in the liquid state at pressure P and temperature T , i.e. V A,mol = V ∗ (P)/n ∗A , and ln x A = ln(1 − x B ) ≈ −x B . One final transformation of this equation is useful. Inside the tube we have xB =
nB nB ≈ . nA + nB nA
(3.60)
In addition Vsolution ≈ V A,mol n A and thus ≈
n B RT . Vsolution
(3.61)
This is the so called van’t Hoff equation (Jacobus van’t Hoff, first Nobel prize in chemistry for his work on chemical dynamics and osmotic pressure, 1901). Note that the osmotic pressure only depends on the molar concentration of component B and temperature (under the approximations we have made in the course of the derivation). Here osmotic pressure is another example for a colligative property. Remark: Reverse Osmosis. According to our derivation leading to Eq. (3.61), it should be possible to apply extra pressure to the tube in Fig. 3.9 and by doing so reduce its solute content. A technical example is desalination of sea water, which is forced through a membrane using a pressure exceeding the osmotic pressure. This process is called reverse osmosis. Example: Osmotic Pressure in Hemoglobin Solutions. As an application we consider the following problem. Use the osmotic pressure data ( pobs. ) from table X in Adair (1928) to estimate the molar mass of (sheep) hemoglobin. Gilbert S. Adair was a pioneer of macromolecular biochemistry and succeeded in determining the correct molecular weight of hemoglobin from osmotic pressure measurements. He also supplied the horse hemoglobin crystals which allowed Max Perutz (Nobel prize in chemistry for his work on the structure of globular proteins, 1962) to obtain the first hemoglobin x-ray structures. We may rewrite van’t Hoff’s equation as RT ≈ . c m Hb Here c = (n H b /V )m H b and m H b is the molar mass of component B (Hb: hemoglobin). Figure 3.10 shows the data from the above reference plotted in the original units. The solid line is a fit on the basis of a theory explained later in this book. Notice first that van’t Hoff’s equation describes the data only at
3.3 Applications Involving Chemical Equilibrium
93
very low hemoglobin concentration. This is expected, because we have used the approximation x A x B . The deviation from van’t Hoff’s equation arise due to non-ideality, which basically means that there are complex solute-solute interactions—something we have no information about at this point. However, we can still determine m H b , i.e. m Hb ≈
RT , /c
in the limit c → 0. From the figure we extract the value /c ≈ 0.3 cm H g/(g/dl). In addition T = 0 ◦ C = 273.15 K. After converting the units, 1cm H g = 1333.224Pa 1g/dl = 10 kg/m3 , we obtain m H b ≈ 57 kg/mol. This is roughly 10 % below the exact value—but not bad at all. The example also shows that van’t Hoff’s equation is valid at small concentrations only. We continue our discussion of osmotic pressure on page 111 and a second time on p. 170 dealing with extensions of different origin. In particular we shall discuss the so called scaled particle theory behind the solid line through the data in Fig. 3.10 beginning on p. 179. This theory allows to estimate the size of Hb (≈5.5 nm in diameter) based on its osmotic pressure data.
-1
/c [cm Hg (g / dl) ]
1.2 1.0 0.8 0.6 0.4 0.2 0.0
0
5
10
15
20
25
30
c [g/dl] Fig. 3.10 Concentration dependent osmotic pressure in hemoglobin solutions
35
94
3 Equilibrium and Stability
Equilibrium Adsorption Consider a gas in contact with a solid surface. Molecules from the gas may adsorb onto and subsequently desorb from the surface. Eventually an equilibrium develops characterized by a constant coverage depending on temperature and the pressure in the bulk gas. Coverage here refers to the net amount of gas adsorbed. We obtain the net amount adsorbed by counting the gas molecules in a column-shaped volume perpendicular to the surface. This column continues out into the bulk gas, where the surface is no longer felt by the gas molecules. Subsequently we subtract the (average) number of gas molecules present in an identical column when the surface is removed (The number is equal to the volume of the column multiplied by the bulk density of the gas). Just how long the column has to be, in order for it to extend into the bulk gas, depends on the interaction forces between the gas molecules and the surface as well as on thermodynamic conditions. In some cases the “interfacial thickness” to good approximation is just one molecular layer. One speaks of monolayer or even sub-monolayer coverage. In other cases the interface is “thicker” and more “diffuse”. The examples in Fig. 3.11 show computer simulation generated gas density profiles above an adsorbing surface at z = 0.5 The units used here are so-called LennardJones units6 —but this is of no particular interest to us at this point. What is shown is the gas number density, ρ(z), as function of distance, z, from the surface. In the top panel we recognize a first peak at z ≈ 1 (cut off at 1) and a second smaller one at z ≈ 2. Beyond the second peak the density levels off (with fluctuations) indicating the bulk phase. The eventual drop at z = 12 is merely due to the finite extend of the simulation box to which the gas is confined. The “gap” between the surface and the first peak is due to the finite extend of the atoms in the surface and the molecules in the gas—ρ is a center of mass number density. This figure shows that at the given conditions there exists a dense layer of adsorbed molecules adjacent to the solid surface. A much less dense second layer is followed by a rather rapid transition to bulk behavior. Altered conditions do change the picture. In this case the temperature is reduced. In fact we approach the saturation line of methane at constant pressure (the transition temperature at this pressure in LJ-units is just about 1). We notice that the adsorbed layer thickness increases as more peaks emerge. However, at this point these graphs merely serve as illustration to bear in mind when we talk about adsorption on solid surfaces.7 An important quantity characterizing the interaction of the molecules with the surface is the isosteric heat of adsorption, qst , defined via
The system is methane gas adsorbing on the graphite basal plane located at z = 0. A computer program generating profiles like these is included in the appendix. The theoretical background needed to understand the program is discussed in Chap. 6. 6 In these units the gas pressure in Fig. 3.11 is P = 0.04. The temperatures from top to bottom are T = 2.0, 1.2, 1.05. 7 We return to Fig. 3.11 in an example starting on p. 206. 5
3.3 Applications Involving Chemical Equilibrium
95
ρ [LJ] 1.0 0.8 0.6 0.4 0.2
0
2
4
6
8
10
12
2
4
6
8
10
12
2
4
6
8
10
12
z [LJ]
ρ [LJ] 1.0 0.8 0.6 0.4 0.2
z [LJ] 0
ρ [LJ] 1.0 0.8 0.6 0.4 0.2
0
z [LJ]
Fig. 3.11 Computer simulation generated gas density profiles above an adsorbing surface at different temperatures
qst = T
∂μs ∂μb −T . ∂ T Vs ,Ns ∂ T Pb
(3.62)
96
3 Equilibrium and Stability
The indices s and b refer to the surface and the bulk, respectively. Note that . . . |Vs ,Ns means “at constant coverage”, whereas . . . | Pb means “at constant (bulk) pressure”. The temperature is the same in both cases. Using the equilibrium condition μs = μb we may write ∂μs ∂μb ∂μb dT = dT + dμs V ,N = d Pb , s s ∂ T Vs ,Ns ∂ T Pb ∂ Pb T
(3.63)
which yields ∂μs ∂μb ∂μb ∂ Pb = . + ∂ T Vs ,Ns ∂ T Pb ∂ Pb T ∂ T Vs ,Ns
(3.64)
Vb /Nb =ρb−1
Combination of this equation with Eq. (3.62) yields another, and perhaps the most common, expression for qst : qst =
T ∂ Pb Pb ∂ ln Pb =− . ρb ∂ T Vs ,Ns ρb T ∂(1/T ) Vs ,Ns
(3.65)
At very low gas pressure one may assume that Ns is proportional to Pb , i.e. Ns = k H Pb + O(Pb2 ).
(3.66)
The leading term is a “surface version” of Henry’s law Eq. (3.46). In this approximation Eq. (3.65) becomes (o)
qst = R
∂ ln k H . ∂(1/T ) Vs ,Ns
(3.67)
(o)
Here qst is the molar isosteric heat of adsorption in the limit of vanishing coverage. Experimentally this quantity may be determined by measuring the amount of adsorbed gas (e.g., by weighing the sample) at a given (low) pressure. The general relation Ns (T, P) vs. P is called adsorption isotherm, the low pressure slope of which again is k H . On p. 206 we return to the isosteric heat of adsorption and discuss one explicit method how to calculate it theoretically.
Law of Mass Action In the following we discuss an important application of Eq. (2.119), i.e. dG T,P ≤ 0.
(3.68)
3.3 Applications Involving Chemical Equilibrium
97
At equilibrium we can use the equal sign and based on Eq. (2.107) (with μdn replaced K μi dn i ) we have by i=1 K
μi (T, P)dn i = 0.
(3.69)
i=1
This equation requires some thought. If K = 1 then Eq. (3.69) implies dn = 0. The inequality Eq. (3.68) applies to cases where, aside from keeping T and P at fixed values, we leave the system alone. In particular we do not change its mass content.8 If K > 1 there exists however the possibility of a suitable relation between the dn i , developed by the system itself, allowing Eq. (3.69) to hold without requiring dn i = 0 ∀ i. For instance we may replace dn i in Eq. (3.69) via dn i = νi dξ,
(3.70)
where of course not all νi do have the same sign. It turns out that chemical reaction equilibria may be described in this fashion. For a chemical reaction the dn i obey according to experimental evidence νi dn i = , dn j νj
(3.71)
where νi and ν j are integers. We proceed writing the chemical potentials of the components as μi (T, P) = μ¯ i (T, P) + RT ln ai .
(3.72)
This particular form is analogous to the special limiting forms Eqs. (3.27) and (3.45). The quantity ai , which is called activity of component i, contains interaction and mixing contributions to the chemical potential of component i, i.e. all effects due to the interactions of this component with all other components. Often the activity is expressed via ai = γi xi ,
8
(3.73)
Potentially this may be disturbing. According to the steps leading from Eq. (2.164) to Eq. (2.165) one may be led to conclude that G = 0 all the time and everything falls apart. However this reasoning confuses two very different situations. Inequality Eq. (3.68) means that we prepare a system subject to certain thermodynamic conditions T and P and leave this system alone until no further change is observed. This fixes the equilibrium value of the free enthalpy, G, for a particular pair T, P. Repeating this procedure for many T, P-pairs we map out the equilibrium values of G above the T -P-plane (cf. Fig. 2.13). With this function G = G(T, P) or G = G(T, P, n) we now can do calculations, differentiating or integrating, involving T , P, n and possibly other variables. This is how we have obtained the Eqs. (2.164) and (2.165). Therefore there is no problem here!
98
3 Equilibrium and Stability
where γi is the activity coefficient. This is the usual terminology in condensed phases. In the gas phase the fugacity, f i = γi P,
(3.74)
where γi is the fugacity coefficient and P is the pressure, replaces ai . We see that the special limiting forms Eqs. (3.27) and (3.45) correspond to γi = 1. The reference chemical potential, μ¯ i (T, P), may be identified if we let γi and xi approach unity. Combining Eqs. (3.69), (3.70), and (3.72) we obtain K
(μ¯ i (T, P) + RT ln ai )νi = 0
(3.75)
i=1
or K
aiνi = K (T, P),
(3.76)
i=1
where K ¯ i (T, P) i=1 νi μ K (T, P) = exp − . RT
(3.77)
Equation (3.76) is called law of mass action and K (T, P), not to be confused with the index K , the number of components, is the equilibrium constant. The equilibrium constant is not really a constant. It depends on T and P. By convention νi < 0 for reactants and νi > 0 for products. The law of mass action in its present form provides little insight. Therefore we study the special case of a gas phase reaction assuming that the gas is ideal. Combining Eqs. (3.26) and (3.21) we obtain (g)
(g)
μi (T, Pi ) = μi (T, Pio ) + RT ln
Pi . Pio
(3.78)
Here Pi is the partial pressure of component i, and Pio is a standard pressure, which remains constant during the reaction. In addition we have (g)
xi
=
Pi , P (g)
(3.79)
cf. (3.28), where P (g) is the total pressure. Combination of Eqs. (3.78 ) and (3.79) yields
3.3 Applications Involving Chemical Equilibrium (g)
99 (g)
(g)
μi (T, Pi ) = μi (T, Pio ) + RT ln
P (g) xi . Pio
(3.80)
Inserting this into Eq. (3.69) we find the law of mass action K (T, P o ) = P i νi
xiνi ,
(3.81)
i
where −RT ln Pio is absorbed into K (T, P o ). Notice that we have omitted the index (g), and we assume that P o = Pio ∀ i. This equilibrium constant is independent of the gas pressure P and the mole fractions xi . Example: A Chemical Reaction. In the following simple example of a chemical reaction, 2H2 + O2 2H2 O,
(3.82)
we have ν H2 = −2, ν O2 = −1, and ν H2 O = 2. Thus Eq. (3.81) becomes K (T, P o ) =
2 1 x H2 O . 2 x P xH 2 O2
(3.83)
Increasing the total pressure (at constant temperature) shifts the reaction equilibrium to the right. Analogously we can see what happens if the concentrations are changed. Remark: What is different if in addition to H2 , O2 , and H2 O another inert gas is present or there is an excess of one or more of the aforementioned components? The corresponding mole fractions do not appear explicitly on the right side of Eq. (3.83), but they do enter into the pressure, P. At this point we may ask: What is a component? Thermodynamic knows nothing about atoms, molecules, and details of the interactions/reactions between them. But we know that there are even smaller building blocks than atoms—electrons, protons, and neutrons. And this is not the end. So what is a component? In principle we may apply thermodynamics on all levels. For the above it is important, however, that there exists a meaningful chemical potential for everything we want to call component. That is a component must exist long enough (on average) under well defined thermodynamic conditions like equilibrium T and P. This requires us to rethink our derivation of the phase rule. Consider the following example for a chemical reaction: 3A A3 .
(3.84)
100
3 Equilibrium and Stability
If we consider A and A3 as components, then the phase rule Eq. (3.12) allows up to four coexisting phases. However, we have an additional equilibrium constraint imposed by Eq. (3.69), reducing the degrees of freedom by one and the maximum number of coexisting phases to three. The modified phase rule therefore is Z = K − Q − + 2 (≥ 0),
(3.85)
where Q is the number of additional constraints imposed via Eq. (3.69). Notice that Q is not necessarily one all the time. There may be independent chemical reactions occurring simultaneously in which case the summation in Eq. (3.69) breaks up into independent parts, e.g. 3A A + A2 A3 . Here we have a system containing three components according to our definition. But for each reaction we have to fulfill Eq. (3.69). Therefore K = 3 and Q = 2. Example: Critical Micelle Concentration. Figure 3.12 shows a sketch of a system containing typo-amphiphilic molecules. Typo-Amphiphilic molecules consist of two covalently bonded moieties—one, depicted as zigzag-line, does not like to be in contact with water, not shown explicitly, whereas the other, depicted as solid circle, does like to be in contact with water. An example of such a molecule is Hexaethylene-glycol-dodecylether (C12 H25 (OC H2 C H2 )6 O H ). In this case it is the C12 H25 -moiety that does not like to be in contact with water. The natural thing to happen therefore is a clustering of the zigzag “tails” into droplets shielded on the outside by their water loving “head” groups. In a sense this is a phase separation, which we study in the next chapter, on a molecular scale. Because of this the molecular “shape” strongly couples to the “shape” of the drop or aggregate and in fact determines it (The aggregates we have in mind can be spherical, cylindrical or transform into layered structures with complicated topology. It also is possible to extend this approach to vesicles. But this is not our topic here.). The type of droplet aggregate we just described is called a micelle. However, our current approach covers other types of aggregates as well. As our starting point we choose the “chemical reaction equation” s A1 As .
(3.86)
Here As denotes a s-aggregate containing s molecules or monomers A1 . We put “chemical reaction” in quotes, because the bonding forces between monomers considered here are different from chemical bonds within molecules. In principle s can be any integer number and therefore Eq. (3.86) represents many
3.3 Applications Involving Chemical Equilibrium
101
“reaction equations”. Expressing this in terms of the chemical potential yields sμ1 = μs .
(3.87)
Assuming low monomer concentration we may use Eq. (3.48), i.e. s μ¯ 1 + s RT ln x1 = μ¯ s + RT ln(xs /s).
(3.88)
Note that the quantity xs /s is the mole fraction s-aggregates. Therefore xs is the mole fraction of monomers in s-aggregates. We may solve for xs , i.e. s x s = s x 1 eα
(3.89)
with α=
1 RT
1 μ¯ 1 − μ¯ s . s
(3.90)
We assume that μ¯ s is an extensive quantity in terms of s and that therefore α is independent of s (see also the next example). Equation (3.89) has an interesting consequence. To see this we note that the total monomer mole fraction is given by x = x1 +
∞ s=m
xs = x1 +
∞ s s x 1 eα .
(3.91)
s=m
Here x1 is the mole fraction of free monomers, whereas the sum is the mole fraction due to all other monomers bonded inside aggregates. We note that m is a minimum aggregate size. In the case of spherical micelles for instance, it accounts for the fact that a certain number of head groups are required to form a closed surface avoiding contact of the tail groups with water. This number may be large—say m ≈ 50—depending of course on the type of monomer. But m = 2 also is possible. This is the case of linear aggregates (chains of monomers—These monomers may be disk-shaped with flexible tails on their perimeter. In water the disk-like cores tend to form stacks. It also is possible to apply this idea to dipolar molecules forming chains due to dipole-dipole interaction.). The right side of Eq. (3.91) is bounded, because x ≤ 1. In particular s sq at q = 1 (geometthis requires x1 eα < 1, because the sum ∞ s=m ∞ diverges s ≈ 4×10−3 if q = 0.8 sq ric series!). Putting in some numbers we find s=50 s −α virtually all of x is due to and ∞ s=50 sq ≈ 3 if q = 0.9, i.e. for x 1 < 0.8e free monomers. Addition of monomers at this point leads to their assembly into
102
3 Equilibrium and Stability
Fig. 3.12 Sketch illustrating the reversible assembly of amphiphilic molecules into micelles
s-aggregate
xs μs monomer s=1
x1 μ1
aggregates. Figure 3.13 illustrates this for different combinations of assumed values for m and α (Notice the change of scale in the third panel.). Because of the sharpness of the “transition” in the typical case of large m the threshold concentration xC MC ≈ e−α
(3.92)
is called critical aggregate concentration or, in the case of micelles, critical micelle concentration (CMC). While the sharpness is governed by m, the amphiphile concentration at which the change of behavior occurs is determined by α. We note that the existence of a CMC is not tied to the specific form of Eq. (3.89). For instance, assuming that monomers may form minimum aggregates only, i.e. only the s = m-term in the sum in Eq. (3.91) is present, still yields a CMC. The true size distribution, xs , in fact is a complicated function of molecular interactions as well as thermodynamic conditions. One interesting and quite general ingredient ignored here is the aggregate dimensionality, which is discussed in the following. More information on molecular assemblies (micelles, membranes, etc.) can be found in Israelachvili (1992) or in Evans and Wennerström (1994).
Surface Effects in Condensation The assumption underlying Eq. (3.90) is that all monomers inside an aggregate are equivalent. For a spherical micelle this is in accord with intuition. But what if we study droplets containing monomers completely embedded in their interior and monomers on their surface? These two are certainly different and Eq. (3.90) no longer holds.
3.3 Applications Involving Chemical Equilibrium
103
1.0
m =5, α =1 0.8
xs
0.6
s= m
0.4
x1
0.2 0.0 0.0
0.2
0.4
0.6
0.8
1.0
x
1.0
m= 50, α = 1 0.8
xs
0.6
s= m
0.4
x1 0.2 0.0 0.0
0.2
0.4
0.6
0.8
1.0
x
0.10
m= 50, α = 4 0.08
xs
0.06
s= m
0.04
x1 0.02
0.00
0.02
0.04
0.06
0.08
0.10
x
Fig. 3.13 Mole fractions free monomers and aggregates vs total monomer mole fraction for different parameter combinations
A simple model for α allowing to distinguish between bulk and surface monomers in aggregates is α = αbulk − δs −1/d ,
(3.93)
104
3 Equilibrium and Stability
where d is the dimensionality of the aggregates (d = 1: linear aggregates; d = 2: disk- or layer-like aggregates; d = 3: spherical aggregates). αbulk is the same for all monomers inside aggregates and independent of s. The second term, −δs −1/d , is a surface contribution. We note that a three-dimensional spherical droplet containing s monomers has a volume proportional to s. Thus its radius is proportional to s 1/3 and its surface is proportional to s 2/3 . Expressed more generally the surface is proportional to s (d−1)/d = ss −1/d . This means that in this simple case μ¯ s can be expressed as μ¯ s = s μ¯ bulk + RT δs (d−1/d) ,
(3.94)
where μ¯ bulk , the chemical potential of a monomer in the interior of an aggregate, is s-independent. What are the consequences? First we study the question whether monomers and/or finite size aggregates can coexist with infinite size aggregates, i.e. the bulk phase. If the answer is yes, then the following must be true: μ¯ bulk = μ¯ bulk + RT δs −1/d +
xs 1 RT ln . s s
(3.95)
This means xs = s exp[−δs (d−1)/d ].
(3.96)
Consequently s xs < 1 is possible only if d > 1. If d = 1 we have xs ∝ s and the sum diverges. This means that the total monomer mole fraction x diverges. The inconsistency that this imposes (note: x ≤ 1) is interpreted as the impossibility of coexistence between monomers and/or finite aggregates with a bulk phase in one dimension! But there is more to discover here. We concentrate on d > 1 and simplify our calculation by requiring the aggregates to be monodisperse, i.e. s is the same for all aggregates. The total free enthalpy therefore is G total = n 1 μ1 +
ns μs . s
(3.97)
Here n s denotes moles monomer on average bound in aggregates. This equation describes coexistence between a gas of monomers and aggregate droplets. Using Eq. (3.94) yields G total = n 1 μ1 +
ns μs,bulk + n s RT δs −1/d , s
(3.98)
where μs,bulk = s μ¯ bulk + RT ln[xs /s]. If we vary the mass distribution between monomers and aggregate droplets near equilibrium we find
3.3 Applications Involving Chemical Equilibrium
105
1 dG total = dn s −μ1 + μs,bulk + RT δs −1/d = 0. s
(3.99)
Note that dn 1 = −dn s . We may usefully apply this equation using ∂μ/∂ P|T = 1/ρ or dμ = ρ −1 d P at constant T . Because the monomers form a gas with density ρgas , while ρliq ρgas is the (liquid) monomer density inside the droplets, we may write −
1 ρgas
+
1 1 ∂s −1/d ≈− ≈ −RT δ . ρliq ρgas ∂P T
(3.100)
Replacing ρgas by P/(RT ), the ideal gas law, and s 1/d by c r , where r is the droplet radius and c is a constant, we find 1 (3.101) d ln P = δd cr or P = P∞ exp
δ . cr
(3.102)
This equation describes the radius, r , of a droplet at equilibrium when the external pressure is P. P∞ is the saturation pressure, when the monomer gas coexists with the infinite bulk phase. Figure 3.14 shows a sketch of P vs. r according to Eq. (3.102). But what happens to droplets which do not have the proper equilibrium radius? At point A in Fig. 3.14 a droplet will find the external pressure too low for its size, which is less than the proper equilibrium size ro , and therefore evaporates monomers. This decreases the radius and the evaporation continues until the droplet disappears. At point B the droplet is under too high a pressure and additional monomers from the gas phase condense on its surface. The droplet continues to grow and finally the limit of a continuous bulk phase is approached. Thus Eq. (3.102) defines the critical size of a droplet at a given pressure. Below the critical size droplets disappear, above the critical size they grow without bound. In turn this means that finite droplets in general are not stable for d > 1. Fig. 3.14 Sketch of P vs. r according to Eq. (4.102) P
B
A P
ro
106
3 Equilibrium and Stability
Debye-Hückel Theory An overall neutral system contains mobile charges. Of special interest in this context are electrolytes, i.e. substances containing free ions. Typically these are ionic solutions, e.g. aqueous solutions of dissociated acids, bases or salts (e.g. 1+ 1− + Cl(aq) ). What we want is an approximate description for N aCl(s) → N a(aq) the electrostatic interaction as part of the chemical potential of the (ionic) charges. On page 55 we had discussed the relations of the free energy and the free enthalpy to the second law. In the present case electrical work must be included and therefore we have dG|T,P ≤ −δwq
(3.103)
or at equilibrium and expressed as free enthalpy per mole. (1.15)
(s)
dμq |T,P = −N A δwq = N A dqφba .
(3.104)
The meaning of last equation on the right in the current context is illustrated in Fig. 3.15. One of the charges, charge q, is shown as thick vertical bar. The charge q is part of a charge density ρ(r ), assumed to be radially symmetric and centered on q. An observer on q should notice that the surrounding charge (distribution) preferentially is negative if q is positive and vice versa. However, this q-induced distribution extends over a finite range only. Beyond a sufficiently large distance r the central charge q is electrically invisible or screened. The quantity δwq is the work done on the system when a infinitesimally small (molar) amount of charge dq is brought in from infinity (index a) and added to the central charge at zero (index b). In principle the result is infinite, no matter how small dq is, and therefore useless to us. But what we really want, is the work due to the screening part of the potential φba -indicated by the index (s)- excluding the “bare” potential, q/r , causing the divergence. It is this screening part of the potential which is the manifestation of the interaction between the charges in the system. (s) But how do we calculate φba ? One equation which comes to mind is Poisson’s equation, i.e. 2 φ( r ) = 4πρ( r ), −∇
(3.105)
where φ( r ) is the electrostatic potential of a charge density ρ( r ). In the present case ρ( r ) = qδ( r) + e
ci z i h i ( r ).
(3.106)
i
The first term on the right is the central charge q at the origin. The second term is the charge density in a volume element δV located at r. The factor e is just the magnitude of the elementary charge. The index i indicates different types of charges
3.3 Applications Involving Chemical Equilibrium
107 ρ (r)
Fig. 3.15 Spatial distribution of charge
q
r
possibly present. ci is the overall number concentration of these charges, and z i is the charging of type i. For instance in the case of N aCl in an aqueous solution there are N a + and Cl − ions for which z N a = +1 and z Cl = −1. But there also may be ions for which z i = ±1, e.g. C O32− with z C O3 = −2 or Z n 2+ with z Z n = +2. The r ) = h i (r ) describes the variation of the i-type screening charge. Notice function h i ( r ) should vanish as r approaches infinity - but this is all we can say about that h i ( r ) at the moment. We therefore anticipate the following form of h i (r ): h i ( r) ez i N A φ( − 1. r ) ≈ exp − h i ( RT
(3.107)
The argument of the exponential is the ratio of the electrostatic energy of one mole of i-charges (at r) divided by the “thermal energy” RT . This form of h i ( r ) is by no means exact. It neglects completely structural correlations between charges in the vicinity of the central charge. It merely considers the effect of the surrounding charges in the form of a smooth “screening field” as part of the potential φ( r ), i.e. no two charges interact directly—each charge interacts with the others through r ) as a measure of the their collective “screening field”. Equation (3.107) relates h i ( probability for finding a certain charge concentration at distance r from the central charge to the electrostatic energy of this assembly. The specific form, however, we can understand only on the basis of microscopic theory as explained in Chap. 5 (notice in particular Eq. (6.143)). Nevertheless, the combination of Eqs. (3.105) to (3.107) yields r) ez i N A φ( 2 − 1 . (3.108) r ) ≈ 4π qδ( r) + e ci z i exp − − ∇ φ( RT i
This is the desired equation for φ( r ). However its nonlinearity is inconvenient and thus we go one step further by expanding the exponential, i.e.
ez i N A φ( ez i N A φ( r) r) exp − ≈1− . RT RT
(3.109)
108
3 Equilibrium and Stability
This additional approximation is quite in line with our above assumption for the form of h( r ). It requires that the electrostatic energy is much less than the thermal energy and thus that the temperature is “high” (the theory still should be applicable at room temperature though). High temperature also tends to diminish structural correlations. The final equation for φ( r ) is 8π e2 N A 2 φ( r ) ≈ 4πqδ( r) − −∇ I φ( r ). RT
(3.110)
The quantity I =
1 2 1 2 ci z i = νi z i c 2 2 i i
(3.111)
=p
is called ionic strength. Here ci = νi c, where c is the electrolyte concentration and νi is the number of j-ions per electrolyte molecule. We solve Eq. (3.110) via Fourier transformation. That is we insert φ( r) =
∞ −∞
i k·r ˆ k)e d 3 k φ(
(3.112)
and δ( r) =
1 (2π )3
∞
−∞
d 3 kei k·r
(3.113)
to obtain ≈ ˆ k) φ(
4πq 1 3 2 (2π ) k + λ−2 D
(3.114)
with λD =
RT . 8π e2 N A I
(3.115)
Insertion of Eq. (3.114) into Eq. (3.112) and solving the integration yields φ( r) ≈ −
q r/λ D e − e−r/λ D . r
(3.116)
3.3 Applications Involving Chemical Equilibrium
109
Because the first term in brackets grows without bound for r → ∞, we discard this unphysical part of the mathematical solution and thus use
φ( r) ≈
qe−r/λ D . r
(3.117)
When the central charge is approached from a distance, its bare potential, q/r , becomes “visible” when r λ D . On the other hand, if r λ D then the potential essentially vanishes, i.e. the central charge is screened. λ D is the Debye screening length. How big is λ D ? We first note that the system of our current units requires the replacement of e2 by e2 /(4π εo εr ) if we want to use SI-units. Notice that εr is the dielectric constant of the background medium containing the charges, e.g. ions in water (water: εr = 78.3 at T = 298 K and P = 1 bar). Thus we have λ D = 1.988 × 10
−3
T εr nm with[T] = K and [c] = mol/l. I
(3.118)
For example, in the case of a 0.1molar aqueous N aCl solution λ D = 0.96nm. The φ (s) we want is obtained by subtracting the bare potential q/r from the above φ( r ), i.e. φ( r )(s) ≈
q qe−r/λ D − . r r
(3.119)
(s) The potential difference φba is
q q(1 − r/λ D − O(r 2 )) − q =− . r →0 r λD
(s)
φba = φ(0)(s) − φ(∞)(s) ≈ lim
(3.120) Returning now to Eq. (3.104) we can write
dμq |T,P ≈ −
N Aq dq. λD
(3.121)
The complete μq we obtain by increasing dq to the full q, i.e.
q
μq ≈ − 0
dq
N Aq 1 N Aq2 =− . λD 2 λD
(3.122)
This is half the electrostatic energy of two charges ±q (SI-units: q 2 → q 2 /(4π εo εr )) at a distance λ D (multiplied by N A ). Notice also that μq is what we must add to
110
3 Equilibrium and Stability
the ideal chemical potential of charge q in order to approximately account for its electrostatic interaction with all other charges in the system. In other words, we may write for the (electrostatic) activity coefficient RT ln γq ≈ −
1 N Aq2 . 2 λD
(3.123)
Before we proceed with an example, we want to discuss the inclusion of excluded volume. Thus far the ions are point-like. We may include the effect of finite ion size as follows. Due to overall neutrality we require − 4πq ≈ −
1 λ2D
∞
d 3r φ(r )
(3.124)
b
(cf. the second term on the right side of Eq. (3.110)). Here b is the radius of the ion carrying the charge q. Inserting φ(r ) = Ar −1 exp[−r/λ D ], where A is a constant, we obtain φ( r) ≈
qe−(r −b)/λ D , r (1 + b/λ D )
(3.125)
instead of Eq. (3.117). The potential difference now becomes (s) φba
= φ(b)
(s)
− φ(∞)
(s)
≈ lim
δr →0
=−
qe−(r −b)/λ D q − r =b+δr r (1 + b/λ D ) r
1 q , λ D 1 + b/λ D
(3.126)
and the charging process yields
μq ≈ −
1 1 N Aq2 . 2 λ D 1 + b/λ D
(3.127)
However, for the moment we continue to use Eq. (3.122), i.e. the limit b → 0, and return to Eq. (3.127) when we discuss the phase behavior of simple systems in Chap. 4. Setting b = 0 results in the so-called Debye-Hückel limiting law.9
9
Peter Debye, Nobel prize in chemistry for his many contributions to the theory of molecular structure and interactions, 1936.
3.3 Applications Involving Chemical Equilibrium
111
Fig. 3.16 Spherical cell with a semipermeable wall containing an electrolyte solution
electrolyte + H2 O T,P +
H2 O
T,P
Example: Osmotic Pressure in Electrolyte Solutions. In this example we begin by studying osmotic pressure from a somewhat different angle than before. The spherical cell in Fig. 3.16 is submerged inside a water (or solvent) reservoir kept at constant temperature, T , and pressure, P. The water passes freely between the cell, which has a constant volume V , and the reservoir. A suitable mechanism allows to add electrolyte (or solute) to the cell. Contrary to the water the electrolyte, which we assume fully dissociated into its ions, cannot pass the cell’s wall. We know from our previous discussion of osmotic pressure that the total pressure inside the cell will rise to P+. The dependence of osmotic pressure, , on solute concentration follows via the Gibbs-Duhem equation (2.168) applied to the interior of the cell, i.e. V d P |T =
n j dμ j |T .
(3.128)
j
Here j stands for the different types of ions (or different solute components). The water chemical potential does not appear, because it may adjust to the same value inside and outside the cell. And the outside water chemical potential is constant of course. According to Eqs. (3.72) and (3.73) we may express the change of the j-ion’s chemical potential via dμ j = RT d ln[x j γ j ].
(3.129)
The desired relation between the osmotic pressure and the electrolyte concentration expressed in moles, n, inside V follows via integration of Eq. (3.128): = id + ex =
RT n RT n n j d ln x j + n j d ln γ j . (3.130) V 0 V 0 j j
112
3 Equilibrium and Stability
Using n j = ν j n the ideal part becomes id
n n RT = νj n d ln . V n H2 O (n ) + j ν j n 0
(3.131)
j
Note that d ln j ν j = 0. We recover the van’t Hoff equation if we replace n H2 O (n) + j ν j n by n ∗H2 O , where n ∗H2 O is the water content of the cell at vanishing electrolyte concentration. This approximation requires a negligible solute content, i.e. small electrolyte concentration, and also neglects compressibility effects. Now we can use d ln[n H2 O (n) + j ν j n] ≈ d ln n ∗H2 O = 0 and thus we find id ≈ v H =
RT ν j n. V
(3.132)
j
Remembering Eq. (3.123), the Debye-Hückel result for the activity coefficient, we can approximate the excess osmotic pressure, ex , via ex ≈ D H =
n RT 1 N Aq j . νj nd − V 2 λD 0
(3.133)
j
With λ−1 D ∝
√ n and
n 0
√ 1 n 1/2 1 nd n = n dn = n 3/2 2 o 3
(3.134)
we finally obtain D H = −
RT . 24π N A λ3D
(3.135)
Of special interest is the ratio φ − 1 ≡ ex /id ≈ D H /v H . Here φ is the osmotic coefficient, i.e. φ ≈1+
D H N A e2 =1− . v H 6RT λ D
(3.136)
Figure 3.17 shows φ vs. electrolyte concentration for Ag N O3 and N aCl in aqueous solution (the data are from Hamer and Wu 1972)—squares: Ag N O3 ; circles:
3.3 Applications Involving Chemical Equilibrium
113
1.00
0.95
0.90
0.001 0.002
0.005 0.010 0.020
0.050 0.100 0.200
c [mol/l] Fig. 3.17 Osmotic coefficient vs. electrolyte concentration
N aCl; solid line: Eq. (3.136). The limiting law is a good approximation at very small electrolyte concentrations only. However, various approximations like the neglect of finite ion size, ion-ion correlations, and the explicit interaction with the solvent quickly cause deviations with increasing electrolyte concentration.
Gibbs-Helmholtz Equation In the following we need the Gibbs-Helmholtz equation: H ∂G/T = − 2. ∂T P T
(3.137)
It may be derived via Eq. (2.105) combined with Eq. (2.111), i.e. ∂G . G = H −T ∂ T P Dividing both sides by T 2 immediately yields Eq. (2.111). It is useful to rewrite Eq. (3.137) in terms of the chemical potential μ j in a multicomponent system. Differentiation of the left side of Eq. (3.137) with respect to n j yields ∂ ∂G/T ∂n j ∂ T P,n i and thus
T,P,n i(= j)
∂ 1 ∂ = G ∂ T T ∂n j T,P,n i(= j)
= P,n i
∂μ j /T ∂ T P,n i
114
3 Equilibrium and Stability
hj ∂μ j (T, P, n i )/T = − 2, ∂T T P,n i
(3.138)
where h j is the partial molar enthalpy of component j. We note that equations analogous to Eqs. (3.137) and (3.138) hold for the free energy, i.e. E ∂ F/T =− 2 ∂ T V T
(3.139)
ej ∂μ j (T, V, n i )/T = − 2, ∂T T V,n i
(3.140)
and
where e j is the partial molar energy of component j. Example: Saha Equation. During the cosmic evolution a phase called recombination occurred. Neutral hydrogen and helium was formed, when the temperature had dropped to about 3000K (In his book Cosmology (Oxford University Press, 2008) Steven Weinberg (Nobel prize in physics for his contributions to the unification of fundamental interactions, 1979) points out that recombination may be misleading because no neutral atoms had ever existed until this point. But this is the usual term and in addition recombination is still occurring today in the atmospheres of stars.). Even though we cannot provide a complete discussion of this process, which can be found in the aforementioned reference, we still want to get a feeling for why it is associated with such a distinct temperature. Here we study the reaction p + e 1s. p and e stand for one proton and one electron, respectively, while 1s denotes the atomic hydrogen ground state. In analogy to the example on p. 99 we write x1s = P K (T, P o ) x p xe assuming ideal gas behavior. However, what we want to calculate is the fraction of ionized hydrogen
3.3 Applications Involving Chemical Equilibrium
X=
115
xp . x p + x1s
We may combine the last two equations into one, i.e. X (1 + S X ) = 1,
(3.141)
where S = (ρ p + ρ1s )P K (T, P o )/ρ. Note that x p = xe and ρi = ρxi , where ρ is the total number density of massive particles in the universe at this time. Equation (3.141) is the Saha equation (Meghnad Saha, 1893–1956, indian astrophysicist). What we really want is X = X (T ) and thus we need the explicit temperature dependence of K (T, P o ). The latter quantity is given by K (T, P o ) =
1 1 o o o μ exp (T, P ) + μ (T, P ) − μ (T, P ) . p e 1s Po RT
Using Eq. (3.138) we have μi (T, P o ) μi (T o , P o ) = − T To
T
dT
To
h i (T ) T 2
.
The partial molar enthalpy is h i = ei + RT , where the internal energy is (o) (o) (o) (o) ei = ei + 3RT /2. We also use e1s − e p − ee = 13.6eV N A , where the right side is the ionization energy for one mole of 1s hydrogen. Overall we obtain (o)
μi (T o , P o ) ei μi (T, P o ) = + T To R
1 1 − o T T
+
5 To ln , 2 T
and thus S = So (ρ p + ρ1s )T −3/2 exp[158000K /T ], where we have used the ideal gas law to replace P and 13.6eV = 1.58 · 105 K . Here So is a number depending on the reference state (T o , P o ), which thermodynamics does not reveal.
116
3 Equilibrium and Stability
This factor we shall obtain later from Statistical Mechanics, where we learn that the chemical potential of an ideal system of point-like particles is μi = (o) RT ln[ρi 3T,i ] + ei . Here T,i =
2π 2 mi k B T (o)
is the so called thermal wavelength and ei is an internal contribution to the particle’s chemical potential in the above sense. Setting the masses of the proton and the 1s hydrogen equal, i.e. m p = m 1s , we find So = 4.14 · 10−22 m 3 . Finally we need to know ρ p + ρ1s . Using the estimate given in Cosmology (chapter 2.3: Recombination and last scattering), i.e. (ρ p + ρ1s )So is roughly 1.75 × 10−24 , we compute X (T ) to obtain the result shown in Fig. 3.18. It is worth noting that neither position nor shape of the step are significantly dependent on the numerical values of ρ p + ρ1s or So (the reader is encouraged to check this). Weinberg points out that the calculation thus far gives the correct order of magnitude of the temperature of the steep decline in fractional ionization, but it is not correct in detail. However, the in depth discussion is complicated and the interested reader is referred to the above reference.
Boiling-point Elevation Figure 3.19 sketches a hypothetical crossing at constant pressure from a liquid phase into the gas in a one-component system consisting of the substance A. The temperature at which this happens is Tb . From what we know already we may guess the Fig. 3.18 Fraction of ionized hydrogen vs temperature
1.0 0.8
X
0.6 0.4 0.2 1.0
1.5
2.0
2.5
T [103 K]
3.0
3.5
4.0
3.3 Applications Involving Chemical Equilibrium Fig. 3.19 Hypothetical crossing of the saturation line
117
P liquid
gas T
Tb
μ
Fig. 3.20 Sketch illustrating boiling-point elevation
liquid
gas T
T
Tb
(i)
(l)
μA
(ii)
T
general form of the attendant chemical potential—in a narrow temperature range around Tb . This guess is shown in the upper portion of Fig. 3.20. The solid line depicts the chemical potential along our path. The dotted lines are extensions of the liquid and gas chemical potentials, respectively, to where they are not stable anymore. Essentially this picture is based on the inequality Eq. (3.68). Now suppose we do add a small amount x B of a second component B to the liquid. According to Eq. (3.45) we find (l)
(l)
(l)
(l)
(l)
δμ A = μ A∗ − μ A = −RT ln x A ≈ RT x B .
(3.142)
118
3 Equilibrium and Stability
As always * indicates the pure A and x A + x B = 1. In addition x A x B . Equation (3.142) predicts a downward shift of the liquid chemical potential of A, which is shown as long-dashed line in Fig. 3.20. We note that we are interested only in the immediate vicinity of the boiling temperature Tb . Therefore this line and the corresponding solid line are parallel to good approximation. Furthermore we assume that the amount of B in the gas phase is negligible or causes only a negligible shift of the gas phase chemical potential. Therefore we find that the intersection of the chemical potentials of A in the liquid phase and the gas phase has shifted to a higher temperature Tb + δT . (l) Using the Gibbs-Helmholtz equation we may relate δμ A at Tb to the boiling-point (l) elevation δT . Geometrically δμ A is given by (l)
δμ A = (ii) − (i),
(3.143)
where (ii) and (i) are defined in the bottom portion of Fig. 3.20. By simple trigonometry (g) ∂μ A∗ (ii) = − δT ∂T
and
Tb
∂μ(l) ∗ A (i) = − δT. ∂T
(3.144)
Tb
Thus we have
(l) δμ A
(g) (l) ∂(μ A∗ − μ A∗ ) ∂μ A∗ =− δT. δT ≡ − ∂T ∂ T Tb
(3.145)
Tb
The Gibbs-Helmholtz equation enters via
(l) δμ A
= =
(l) (3.137)
δμ A
=
μ A∗ ∂μ A∗ ∂ T − δT = − δT ∂ T Tb ∂T T Tb μ A∗ (Tb ) ∂μ A∗ /T − +T δT Tb ∂T Tb vap h δT. Tb
(3.146)
Notice that μ A∗ (Tb ) = 0 (chemical equilibrium!) and vap h is the molar enthalpy change upon crossing from pure liquid A to pure gaseous A, i.e. the enthalpy of vaporization of pure A. The enthalpy change during a phase transition also is called latent heat—here latent heat of vaporization. Table 3.2 compiles latent heats of vaporiza-
3.3 Applications Involving Chemical Equilibrium
119
Table 3.2 Latent heats of vaporization and melting for various compounds Compound Ice Water N2 O2 Octane
vap H [J/g]
melt H [J/g]
2838(T = 273.15K ) 2258(T = 373.12K ) 199(T = 77.35K ) 213(T = 90.2K ) 364(T = 298K )
333.6(T = 273.15K ) 25.3(T = 63.15K ) 13.7(T = 54.36K ) 182(T = 273.5K )
tion and melting for a number of substances. We remark in this context that the heat T content of a substance build up without changing phase, H = m T12 dT C p (T ), where m is the mass of the substance, is called sensible heat. Combining Eq. (3.146) with Eq. (3.142) we finally arrive at δT ≈
RTb2 (l) x . vap h B
(3.147)
If we look up vap h for the transition of water to steam at 1bar , e.g. from Table 3.2, (l) we obtain δ(T /K ) ≈ 28.5x B , i.e. for amounts B in accord with our above approximations the shift is quite small.
Freezing-point Depression Here the above solution containing mostly A and little B is in equilibrium with the solid A, where again the B-content is negligible. Analogously to Fig. 3.19 we may draw the sketch shown in Fig. 3.21. Apparently this time the transition temperature, which is the melting temperature Tm , is reduced by the addition of B. A completely analogous calculation yields Fig. 3.21 Sketch illustrating freezing-point depression
μ
solid liquid
T
Tm
T
120
3 Equilibrium and Stability
T [K] 0.01
0.02
0.03
x
-0.5 -1.0 -1.5 -2.0 -2.5
--3.0 Fig. 3.22 Theoretical predictions of freezing point depression compared to experimental data
δT ≈ −
RTm2 (l) x , melt h B
(3.148)
where melt h is the molar melting enthalpy at given pressure. For water at 1bar (l) we obtain the freezing-point depression δ(T /K ) ≈ −103x B . Figure 3.22 shows this relation (solid line) in comparison to data points for D-Fructose (crosses) and Silvernitrate (Ag N O3 ) (open squares) taken from HCP. Notice that in the case of Ag N O3 the mole fraction refers to mole ions, i.e. Ag + and N O3− taken individually. The dashed line, apparently an improved description of the Ag N O3 -data, is discussed in the next section. Remark: Cooling. This is a good place to address the following question. Figure 3.23 depicts a glass of water with a floating ice cube. Their respective weights are 200 and 20 g and their momentary temperatures are 20 and 0 ◦ C. What will the temperature of the contents of the glass be after the ice has melted? We assume, as so often, no transfer of heat between the contents of the glass and its surroundings including the glass itself. Then according to the first law we Fig. 3.23 A glass of liquid water with a floating ice cube
20 g ice T=0°C 200g liquid water T=20°C
3.3 Applications Involving Chemical Equilibrium
121
have E + w = E 220g − E 200g − E 20g + P(V220g − V200g − V20g ) = 0, (3.149)
i.e. the overall enthalpy change is zero: H220g − H200g − H20g = 0.
(3.150)
Here the indices refer to the initial water and ice by their respective weights as well as to the final liquid by its weight. Neglecting for the moment the melting enthalpy of the ice, i.e. the enthalpy change when the ice is converted into liquid at T = 0 ◦ C, we obtain the final temperature T via 220gC P T − 200gC P T20 ◦ C − 20gC P T0 ◦ C = 0.
(3.151)
C P is the isobaric heat capacity (we use its value at T = 0 ◦ C which is 4.22 J/(gK)), also assumed to be constant in the relevant temperature range. The resulting final temperature after the ice has melted is T = 18 ◦ C. But this is incorrect, because we have not yet included the enthalpy change during melting. There is a cost associated with the breaking down of the ice structure—most notably the reduction in hydrogen bonding. The price is paid in the form of heat extracted from the content of the glass. This melting enthalpy is tabulated for numerous substances in HCP, where we find melt H = 6.01 kJ/mol for water at ambient pressure and 0 ◦ C. Including this contribution yields (20 g/18g)melt H + 220 gC P T − 200 gC P T20 ◦ C − 20 gC P T0 ◦ C = 0. (3.152) Here 18g is the molar weight of water. The new result is T = 11 ◦ C. This is considerably colder than the previous temperature. Apparently the transition enthalpy is the major contribution10 ! If we redo our calculation with more ice, let’s say 100 g, we obtain even lower temperatures—T = −13 ◦ C in this case. We immediately object that this is unreasonable, because the whole content of the glass freezes before reaching this temperature. Correct11 ! Nevertheless it brings up an idea connecting our present discussion to freezing point depression. We need to depress the freezing point sufficiently in order to reach such low temperatures. 10
This also is something to keep in mind when buying a new washing machine. A higher spin speed is usually better, because of the decreased residual water content in the laundry. This water must be evaporated in the dryer, and the enthalpy of evaporation again is considerable. On the other hand, a modern condenser dryer often is capable of reclaiming some of the invested energy upon condensation. 11 More precisely, the process comes to a halt at coexistence of ice and liquid water.
122
3 Equilibrium and Stability
From Fig. 3.21 we can see that melting above Tm results because the chemical potential of the liquid is lower. Therefore melting continues until the temperature is Tm .12 If for instance we add salt to our liquid water in the above example, we may, depending on the amount we add, lower the temperature far below the freezing temperature of pure water. In addition we may take advantage of a second and third enthalpy change associated with a possible phase change of the substance we add or the mixing process itself. If we select the proper substances then we can get quite low temperatures in this fashion—around −100 ◦ C! Of course, the disadvantage is that this method allows to maintain low temperatures over short periods of time only. Remark: The above reasoning is very much simplified. It is incorrect to conclude that increasing the amount of solute (e.g., salt) allows to continuously depress the freezing point. For instance in the case of N aCl we can only get down to about −21.1 ◦ C. At lower temperatures ice and solid salt coexist. What this means we discuss in the next chapter, where we study simple phase diagrams (in particular liquid-solid coexistence in binary systems starting on p. 160).
The Osmotic Coefficient Revisited Boiling point elevation and freezing point depression can be tied to the osmotic coefficient, φ, and are practical means for its measurement. We start with the GibbsDuhem equation at constant pressure and temperature: − n A dμ A
T,P
=
n j dμ j
j
T,P
.
(3.153)
Here j stand for different solute components. This equation expresses an infinitesimal change of the A-chemical potential via corresponding changes of the j-chemical potentials. The right side of this equation is comparable to the right side of Eq. (3.128). The difference is that in Eq. (3.153) the pressure is constant while in Eq. (3.128) it is not. However, in the liquid phase, where the compressibility is small, we may to very good approximation equate the two right sides and consequently we arrive at δμ A
T,P
≈−
V id φ. nA
(3.154)
Here δμ A is the chemical potential change due to increasing the solute concentration from zero to some final concentration. The sign is just the opposite of the definition in Eq. (3.142) and thus
12
Under “pool conditions” the sun transfers heat to the contents of our glass and the melting continues until the ice is gone.
3.3 Applications Involving Chemical Equilibrium
123
(l)
n vap h φ ≈ A δT. 2 j n j RTb
(3.155)
This is the desired equation for the osmotic coefficient in terms of the boiling-point elevation, where we have inserted the van’t Hoff equation for id . An analogous relation follows for the osmotic coefficient in terms of the freezing point depression: (l)
n melt h φ ≈ − A δT. 2 j n j RTm
(3.156)
The dashed line in Fig. 3.22 is obtained if the osmotic coefficient is calculated via Debye-Hückel theory according to Eq. (3.136). However, the reader should be aware that Ag N O3 is an example for which the limiting law works particularly well.
Chapter 4
Simple Phase Diagrams
4.1 Van Der Waals Theory 4.1.1 The Van Der Waals Equation of State The van der Waals1 theory assumes a molecular structure of matter, where matter means gases or liquids. The interaction between molecules requires modification of the ideal gas law: V = n RT, P P+a(n/V )2
i.e. P=
V −nb
n 2 n RT −a . V − nb V
(4.1)
Molecules, or atoms in the case of noble gases, at close proximity tend to repel. The attending volume reduction is −nb, where b is a parameter accounting for the exclusion of (molar) volume that a particle imposes on the other particles in V . At large distances particles attract, which in turn reduces the pressure. The particular form of this pressure correction, i.e. a(n/V )2 , may be motivated as follows. The number of particle pairs in a system consisting of N particles is N (N −1)/2 ≈ N 2 /2. Expressed in moles this leads to the factor n 2 . The attraction is limited to “not too large” particle-to-particle separation. We assume that two particles feel attracted if they are in the same volume element V . The probability that two particular particles are found within V simultaneously (V /V )2 . Assuming this to be true for all possible pairs leads to an overall number of attracted molecules proportional to (n/V )2 . The resulting Eq. (4.1) is the van der Waals equation of state for gases and 1
Johannes Diderik van der Waals, Nobel prize in physics for his work on the phase behavior of gases and liquids, 1910. R. Hentschke, Thermodynamics, Undergraduate Lecture Notes in Physics, DOI: 10.1007/978-3-642-36711-3_4, © Springer-Verlag Berlin Heidelberg 2014
125
126
4 Simple Phase Diagrams
liquids. The (positive) parameters a and b are characteristic for the specific material. The van der Waals equation of state is by no means accurate, but its combination of simplicity and utility is outstanding. The parameters a and b may be estimated by measuring the pressure as function of temperature at low densities. The result may then be approximated using the following low density expansion of Eq. (4.1): P = RT
∞
Bi (T ; a, b)
n i
i=1
V
,
(4.2)
where B1 (T ; a, b) = 1
(4.3)
a B2 (T ; a, b) = b − RT B3 (T ; a, b) = b2 .. .
(4.4) (4.5)
(ex p)
(T ) by fitting low are so called virial coefficients. In practice one determines B2 order polynomials to experimental pressure isotherms at low densities. The resulting (ex p) (T ) is then plotted versus temperature. Now a and b may be obtained by fitting B2 Eq. (4.4) to these data points. If we introduce the following reduced quantities, p, t, and v, via P = Pc p T = Tc t
(4.6) (4.7)
V = Vc v,
(4.8)
where 1 a 27 b2 8 a RTc = 27 b Vc = 3nb, Pc =
(4.9) (4.10) (4.11)
we may rewrite the van der Waals Eq. (4.1) into p=
3 8t − . 3v − 1 v 2
(4.12)
This is the so called universal van der Waals equation. It is universal in the sense that it does no longer depend on the material parameters a and b. Notice that the ideal
4.1 Van Der Waals Theory
127
gas law in these units is pid.gas =
8t . 3v
(4.13)
4.1.2 Gas-liquid Phase Transition The upper portion of Fig. 4.1 shows plots of the universal van der Waals equation for three different values of t (solid lines). Of course Eq. (4.12) always deviates from the ideal gas law at low v. In fact we did not plot the pressure for v-values below the singularity at v = 1/3, because there the molecules overlap. But we also notice that the universal van der Waals equation exhibits strange behavior if t < 1.0. There is a v-range in which the pressure rises even though the volume increases. Here we find an isothermal compressibility κT < 0—in clear violation of the mechanical stability condition in Eq. (3.16)! Had we plotted Eq. (4.12) for even smaller t-values, we would have obtained negative pressures in addition. All in all, for certain v and t, the van der Waals equation does describe states which cannot be equilibrium states. It turns out however that we can fix this problem, and at the same time we may describe a new phenomenon—the phase transformation between gas and liquid. To understand how the model may be fixed we look at the free energy obtained via integration of the pressure F(V ) = Fo −
V
d V P,
(4.14)
Vo
cf. (2.109). The bottom part of Fig. 4.1 shows the result obtained using the universal van der Waals equation (with Fo = 3) for the same three temperatures as above. The t = 0.9-curve, which violates mechanical stability according to the attendant pressure isotherm, is sketched in somewhat exaggerated fashion in Fig. 4.2. We notice that the system represented by the filled black circle may lower its free energy by decomposing into regions in which the free energy is fl or f g . In between the free energy is f = fl x + f g (1 − x)
with
x=
v − vg . vl − vg
(4.15)
Notice that this is the lowest free energy the system can achieve via decomposition into regions with high density, denoted vl , and regions with low density, denoted vg . The respective volume fractions of the two different regions are assigned by the parameter x (note: v = xvl + (1 − x)vg ) according to the value of v. In other words, for volumes vl < v < vg the homogeneous system is unstable relative to the decomposed or inhomogeneous system. Imagine we move along an isotherm t < 1 starting from a large volume v > vg . We are in a homogeneous so called gas phase. Upon decreasing v we are entering the range vg > v > vl . Here, depending on the value of v, we observe a “mixture”
128
4 Simple Phase Diagrams p
2
1 t =1.1 t =1.0 t =0.9
0
v 0.5
1.0
1.5
2.0
vg
2.5
3.0
vl
f -1
-2
t =1.1
t =1.0
t = 0.9
Fig. 4.1 Van der Waals pressure and free energy vs volume at three different temperatures
of regions having a homogeneous density n/vg or n/vl . As we approach vl the volume fraction of the latter regions increases to unity. If vl ≥ v we are again inside a homogeneous system—the liquid phase. This augmented van der Waals theory therefore predicts a phase change from gas to liquid and vice versa. Notice also that for vg ≥ v ≥ vl
f g − fl ∂ f
p =− =− = constant ∂v T vg − vl
(the straight line in Fig. (4.2) is the common tangent to f at vl and vg ) and f g − fl = p vl − p vg
or
f g + p vg = fl + p vl , =nμg
=nμl
4.1 Van Der Waals Theory
129
Fig. 4.2 Reduction of the van der Waals free energy via phase separation below t = 1
f
fl vl
vg f'
f
fg
v
which means that mechanical and chemical stability are satisfied. In turn we may calculate vl and vg via the conditions p(t, vl ) = p(t, vg ) and μ(t, vl ) = μ(t, vg ) based on the universal van der Waals equation itself, i.e.
−
vl vo
p(t, vl ) = p(t, vg ) and vg dvp(t, v) + p(t, vl )vl = − dvp(t, v) + p(t, vg )vg vo
(4.16)
(4.17)
using nμ = f + pv. The numerically obtained values vl (t) and vg (t) are shown as a dashed line, the binodal line, in the upper part of Fig. 4.1. The area beneath the binodal line is the gas-liquid coexistence region. Notice that no solutions exist if t > 1, i.e. no gas-liquid phase transition is encountered above t = 1. In addition to the binodal line there is a dotted line, the spinodal line, which indicates the (mechanical) stability limit. This means that the isothermal compressibility, κT , is negative below this line. We remark that, among other methods, the simultaneous numerical solution of Eqs. (4.16) and (4.17) may be programmed as “graphical” search for the intersection of pressure and chemical potential in a plot of μ(v) versus p(v). An example is shown in Fig. 4.3 for t = 0.9. The largest t-value for which a solution is obtained is t = 1. Here one finds vl = vg . The corresponding values of the unreduced pressure, temperature, and volume are Pc , Tc , and Vc given by the Eqs. (4.6–4.8). We can find this so called critical point directly by simultaneous solution of Pc = P(Tc , Vc ), d P/d V |Tc ,Vc = 0, and d 2 P/d V 2 |Tc ,Vc = 0 (the second and third equations are due to the constant pressure in the coexistence region). We may rewrite Eq. (4.17) using p (t) = p(t, vl ) = p(t, vg ) as
130
4 Simple Phase Diagrams p (v) 0.72 0.70 0.68 0.66 0.64 0.62 0.60
(v) -4.20
- 4.15
-4.10
-4.05
- 4.00
- 3.95
Fig. 4.3 Pressure vs chemical potential below t = 1
vg
vl
dvp(t, v) − p (t)(vg − vl ) = 0.
(4.18)
The left side of this equation is the sum of the two shaded areas, one positive and one negative, in the upper graph in Fig. 4.1. The equation states that the two areas between the van der Waals pressure p(t, v), and the constant pressure, p (t), which replaces it between vg and vl , are equal. Therefore vg and vl may also be found graphically via this equal area or Maxwell construction . We remark that between vg and vl the van der Waals pressure isotherms are said to exhibit a van der Waals loop. Figure 4.4 shows the possibly simplest of all phase diagrams in the t-v- and in the t- p-plane. The solid line in the upper diagram is the same as the dashed line, i.e. the phase coexistence curve, in Fig. 4.1. The dotted line is the spinodal. The lower graph shows the phase boundary between gas and liquid in the pressure-temperature plane. Notice that here no coexistence region appears because the pressure is constant throughout this region (at constant t). The crosses are vapor pressure data for water taken from HCP. The next figure, Fig. 4.5, shows three isobars above, at, and below the critical pressure. Figure 4.6 shows the van der Waals chemical potential along three isobars close to the critical point (top) and over an expanded temperature range (bottom), i.e. we compute μ along three horizontal lines in the lower panel of Fig. 4.4 just below, at, and just above the critical pressure. For p = 0.96 we cut across the gas-liquid phase transition. Notice that the dashed lines indicate the continuation of the liquid (low t) and gas (high t) chemical potentials into their respective metastable region, i.e. the region between spinodal and binodal line. This justifies our sketches of the chemical potential in Fig. (3.20) and in principle also in Fig. (3.21). At p = 1 the chemical potential still exhibits a kink, whereas above its slope changes smoothly.
4.1 Van Der Waals Theory
131
t 1.0 0.9
critical point
0.8
liquid
gas
0.7 0.6 0.5 0.4
0.5
1.0
5.0
v 50.0 100.0
10.0
p 1.0
critical point
0.8 0.6 liquid
0.4
gas
0.2
t
0.0 0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
Fig. 4.4 Phase diagrams in the t-v- and in the t-p-plane 1/v 3.0 2.5 2.0
p = 0.5 1.5
p =1.0
1.0
p =1.5
0.5
0.6
0.8
1.0
Fig. 4.5 Isobars in the vicinity of the critical pressure
1.2
1.4
t
132
4 Simple Phase Diagrams 3.80 p 1.04 3.82 3.84 p 1.0 3.86 p 0.96 3.88 3.90 0.985
0.990
0.995
1.000
t
1.005
1.010
3.80
p 1.04
3.82 p 1.0
3.84 3.86 p 0.96
3.88 3.90 0.90
t 0.92
0.94
0.96
0.98
1.00
Fig. 4.6 Van der Waals chemical potential along three isobars
The general quality of the van der Waals equation is nicely demonstrated in Fig. 4.7.2 The figure shows coexistence data for seven different substances plotted in units of their critical parameters. The data, in almost all cases, indeed fall onto a universal curve. This behavior is called law of corresponding states. The universal van der Waals equation certainly is not an exact description, but considering its simplicity the agreement with the experimental data is quite remarkable! We briefly return to the second virial coefficient, B2 (T ; a, b), in Eq. (4.4). Figure 4.8 illustrates the comparison between the van der Waals prediction (solid line), i.e. 27 n B2 1 1− , (4.19) = Vc 3 8t and experimental data from HTTD (Appendix C). The agreement with the experimental data is qualitatively correct. We note however, that the form of B3 in Eq. (4.5) 2
The data shown here are taken from the book by Stanley (1971); the original source is Guggenheim (1945).
4.1 Van Der Waals Theory
133
T Tc 1.0
gas
liquid
gas liquid coexistence
0.9
Ne N2 Xe CO CH4 O2 Kr
0.8 0.7 0.6
ρ ρC 0.0
0.5
1.0
1.5
2.0
2.5
3.0
Fig. 4.7 Law of corresponding states demonstrated for various compounds in comparison to the van der Waals prediction n B2 Vc 1
2
3
4
5
6
t
0.5 CH4 O2 Ar
1.0 1.5 2.0 2.5 3.0
Fig. 4.8 Reduced second virial coefficient according to the van der Waals equation in comparison to experimental data
is an oversimplification. The third virial coefficient, B3 , is not independent of temperature as this equation suggests. Notice that B2 (T ) = 0 defines the so called Boyle temperature, TBolyle . According to the van der Waals theory TBoyle =
27 Tc . 8
(4.20)
Another quantity of interest is the compressibility factor at criticality, for which the van der Waals theory predicts that it is a simple universal number: 3 Pc Vc = = 0.375. n RTc 8
(4.21)
In the cases of argon, methane, and oxygen the experimental values are close to 0.29.
134
4 Simple Phase Diagrams
4.1.3 Other Results of the Van Der Waals Theory The gas-liquid phase behavior described by the van der Waals theory is considered “simple”. In this sense it is a reference distinguishing “simple” from “complex”. We emphasize that this refers to the qualitative description rather than to the quantitative prediction of fluid properties. Other phenomenological equations of state may be better in this respect, but it is the physical insight which here is important to us. Because of this we compute a number of other thermophysical quantities in terms of t, p, and/or v.
Isobaric Thermal Expansion Coefficient Figure 4.9 shows the temperature dependence of the isobaric thermal expansion coefficient, α p , below, at, and above the critical pressure. The dashed line is the ideal gas result. Below the critical pressure a jump occurs when the gas-liquid saturation line (cf. Fig. 4.4; right panel) is crossed. At the critical point we observe a divergence. Above the critical point a maximum marks the smooth “continuation” of the gas-liquid saturation line, which sometimes is called Widom line.
Isothermal Compressibility Figure 4.10 shows the volume dependence of the isothermal compressibility, κT , below, at, and above the critical temperature. The dashed lines are the continuation of κT into the metastable region. Notice that κT diverges when ∂ p/∂v|T = 0. This condition defines the stability limit, κT ≥ 0, i.e. inside the gap between the dashed lines the van der Waals equation gives negative κT . Notice that κT also diverges at the critical temperature. Above the critical temperature κT exhibits a maximum near αp 10
p 1.0
8 p 0.5 6
4
p 1.5
2
t 0.6
0.8
1.0
1.2
1.4
Fig. 4.9 Temperature dependence of the isobaric thermal expansion coefficient for different pressures
4.1 Van Der Waals Theory
135
T
10
t 1.0
8
6
t 0.98
4 t 1.02 2
t 1.4
v
0
0
1
2
3
4
Fig. 4.10 Volume dependence of the isothermal compressibility in the vicinity of the critical temperature
the critical volume, which diminishes as the temperature increases. In general the compressibility increases as v increases—the gas is less dense and easier to compress.
Isochoric Heat Capacity Having discussed α P and κT the next obvious function to look at is the isochoric heat capacity, C V . It turns out that within the van der Waals theory all we obtain is C Vvd W = C Vvd W (T ),
(4.22)
i.e. C Vvd W is a function of temperature only. We show this via ∂ ∂ F
∂ ∂ F
=
. V T V ∂ V ∂ T V T ∂ T ∂
=− ∂∂ TP
=− ∂∂VS
V
T
Therefore ∂ 2 P
∂ ∂ S
∂ ∂ S
= =
∂T 2 V ∂T ∂V T V ∂V ∂T V T
(2.146)
=
∂ C V
1 ∂C V
=
. ∂V T T T ∂V T
This proves Eq. (4.22), because according to the van der Waals equation ∂ 2 P/∂ T 2 |V = 0. Remark: The above equation, i.e. ∂C V
∂ 2 P
=T
, ∂V T ∂T 2 V
(4.23)
136
4 Simple Phase Diagrams
may be integrated to yield 2 ∂ B2 (T ) 2 ∂ B2 (T ) n +T , ≈ −n R 2T ∂T ∂T 2 V
C V − C V,ideal
(4.24)
which is a low density approximation to C V . Of course this correction vanishes if the second virial coefficient, B2 (T ), of the van der Waals theory, cf. (4.4) is used.
Inversion Temperature On p. 51 we had discussed the Joule-Thomson coefficient. Based on the universal van der Waals equation (4.12) we want to calculate the inversion line in the t- p-plane. Equation (2.94) is inconvenient, because we have to express v in terms of t and p. However, using Eq. (A.2) we may write
∂p
∂t
∂v
∂v ∂ p
=−
= −
v . ∂p ∂t p ∂ p t ∂t v
∂v t
The inversion line now is the solution of 0=t
∂ p
∂ p
+ v, ∂t v ∂v t
which we may find analytically: √ √ p = 12( t − 3/4)( 27/4 − t).
(4.25)
The inversion temperatures at p = 0 therefore are tmin = 3/4 and tmax = 27/4. Equation (4.25) is shown in Fig. 4.11. The curve encloses the area in the t- p-plane, where the Joule-Thomson coefficient, μ J T , is positive (cooling). Outside this area μ J T is negative corresponding to heating. Experimental data for methane, oxygen, and argon from Fig. 4.12 in Hendricks et al. (1972) are included for comparison. Qualitatively the van der Waals predictions are correct. But the quantitative quality is quite poor for t- p-conditions far from the critical point (the gas-liquid saturation line (cf. Fig. 4.4; right panel) terminating in the critical point (circle) is included), which we use to tie our theory to reality. It is interesting to work out ∂v/∂t| p based on the virial expansion Eq. (4.2), because this allows a better understanding of the Joule-Thomson effect on the basis of molecular interaction. To leading order we find ∂(b2 /t) ∂v
v ,
= +t ∂t p t ∂t
(4.26)
4.1 Van Der Waals Theory
137
p 12 CH4 O2 Ar
10 8
JT
0 heating
6 4
JT
0 cooling
2 0
t 0
1
2
3
4
5
6
7
Fig. 4.11 Inversion temperature according to Eq. (4.25) including experimental data
b
a b
p liquid
t a c
c
p a
gas t=1
liquid
a
liquid
a
gas c
coex
c a gas
coex
v
v
t
Fig. 4.12 Different paths approaching the gas-liquid critical point
where b2 = n B2 (T )/Vc . Consequently the Joule-Thomson coefficient in this approximation is Vc 2 ∂(b2 /t) . (4.27) t μJ T = CP ∂t This equation is general, i.e. we have not yet used the van der Waals equation of state. If we do this, i.e. we insert Eq. (4.4), the result is Vc−1 C P μ J T
1 = 3
27 −1 . 4t
(4.28)
We recognize that the equation describes the Joule-Thomson coefficient at P = 0 (or small P) near the upper inversion temperature. In particular we verify that μ J T > 0 for t < tmax and μ J T < 0 for t > tmax .
138
4 Simple Phase Diagrams
Van Der Waals Critical Exponents Close to the gas-liquid critical point one can show that the quantities ±δp, ±δv, and ±δt, which are small deviations from the critical point in terms of the variables p, v, and t, are simply related to each other as well as to thermodynamic functions like the isobaric thermal expansion coefficient, α p , and the isothermal compressibility, κt (as well as all others!).3 Again we focus on the universal van der Waals equation, i.e. Eq. (4.12). Along the critical isotherm, path (a) in Fig. 4.12, we set p = 1 + δp and v = 1 − δv. Inserting this into Eq. (4.12) we obtain 3 δp = − δv 3 + O(δv 4 ). 2 Ignoring constant factors and additional terms, like the corrections to the leading behavior, we write instead δp ∼ ±δv 3
where ± : v = 1 ∓ δv.
(4.29)
Approaching the critical point from above, t = 1 + δt, along the critical isochor, v = 1, i.e. path (b) in Fig. 4.12, we find δp ∼ δt.
(4.30)
Another special line is the coexistence curve, shown as the dashed line in Fig. 4.1. On the sketches in Fig. 4.12 the path along the coexistence curve is labeled (c). We insert t = 1 − δt and v = 1 ± δv into the universal van der Waals equation and expand the result in powers of δv. Finally we assume δv ≡ δv± = c± δt β .4 Note that δt and δv are positive. The result is 3 3 3β p± = 1 − 4δt ∓ c± δt ± 6c± δt 1+β + . . . . 2 Here . . . stands for higher order terms O(δt 4β ) and O(δt 2+β ), and + and − stand for the gas and the liquid side of the coexistence curve respectively. The stability condition Eq. (4.16) requires p− = p+ . Setting c+ = −c− fulfills this equality but does not yield the desired result because v− = v+ . The only other solution requires 3 ± 6c = 0. Consequently we obtain 3β = 1 + β and ∓ 23 c± ± β=
1 2
and
c± = 2,
(4.31)
The small quantities δp, δv, and δt are all positive. This is the leading term in a power series expansion of δv in δt. Notice that the coexistence curve is not symmetric with respect to reflection across the critical isochore—except very close to the critical point (cf. below). 3 4
4.1 Van Der Waals Theory
139
i.e. the leading relation between δv and δt along the coexistence curve is δv ∼ δt 1/2 .
(4.32)
We may use this to work out the dependence of the isothermal compressibility, κt = −(1/v)∂v/∂ p|t , near the critical point and along the coexistence curve. Again we insert t = 1 − δt and v = 1 ± δv into the universal van der Waals equation and expand the result in powers of δv. We then work out the derivative ∂ p/∂v|t and insert the above result Eq. (4.32). This yields κt−1 ∼ δt.
(4.33)
∂v
∂t p
= − ∂∂tp / ∂∂vp , cf. (A.3) we obtain
α p ∼ δt −1 .
(4.34)
Using the general thermodynamic relation
v
t
the δt-dependence of the isobaric expansion coefficient, α p = (1/v)∂v/∂t| p , near the critical temperature and also along the coexistence curve. Because ∂ p/∂t v to leading order contributes a constant only, we obtain, as above for κt ,
We return briefly to the critical isochore, v = 1, and compute the δt-dependence of κt−1 , when we approach the critical point via this path. Working out ∂ p/∂v|t and setting v = 1 as well as t = 1 + δt yields κt−1 ∼ δt
(4.35)
as before. Only the prefactors (called scaling amplitudes) are different, i.e. κt−1 ≈ 12δt on the path along the coexistence curve and κt−1 ≈ 6δt along the isochore.
Again we find the relation Eq. (4.34) for the same reason as before, i.e. ∂ p/∂t v to leading order contributes a constant only. The reader may want to work out the divergences of α p in Fig. 4.9 and κt in Fig. 4.10 to leading order in δt and δv, respectively. The result is α p ∼ δt −2/3 along the critical isobar and κt ∼ δv −2 along the critical isotherm. The exponents in the above power laws are called critical exponents. Table 4.1 compiles a selected number of them together with their definition, thermodynamic conditions, and van der Waals values. Here ρl − ρg is the density difference across the coexistence curve.5 This quantity is called order parameter. Notice that by construction the order parameter vanishes above Tc . In addition, δT = |T − Tc | and δ P = |P − Pc |. Notice also that we have not yet talked about the heat capacity exponent α.6 The prime indicates the same critical exponent below Tc . The van der Waals theory yields the same values for the two exponents listed here, i.e. α = α What is the relation between ρl − ρg and ±δv? Setting ρl = ρc + δρl and ρg = ρc + δρg and using |δρ/ρ| = |δv/v| yields ρl − ρg ∝ δv− + δv+ . 6 The present critical exponent notation is standard. We want to adhere to it, even though certain letters are used for other quantities also. 5
140
4 Simple Phase Diagrams
Table 4.1 Selected critical exponents and their vdW-values Exponent
Definition
α α β γ γ
C V ∼ δT α C V ∼ δT α ρl − ρg ∼ δT β κT −1 ∼ δT γ κT −1 ∼ δT γ
δ
δ P ∼ ±(ρl − ρg )δ
Conditions
vdW-value
V = Vc (T < Tc ) V = Vc (T > Tc ) coex curve coex curve (T < Tc ) V = Vc (T > Tc ) +: ρ > ρ c T = Tc -: ρ < ρc
0 0 1 2
1 1 3
and γ = γ . But the van der Waals values are not correct! Even though the correct exponent values turn out to be nevertheless the same below and above Tc , we again adhere to the (safe) standard notation, which distinguishes the two conditions. α P does not appear in this list, because, as we have seen, its exponents are also γ and γ at the indicated conditions.
4.2 Beyond Van Der Waals Theory Thermodynamic Scaling The preceding discussion of van der Waals critical exponents has highlighted the power law-relations connecting thermodynamic functions close to the gas-liquid critical point. This led to the idea to express the latter as generalized homogeneous functions.7 That is if f = f (x, y) is a generalized homogeneous function then f (x, y) = λ−1 f (λ p x, λq y),
(4.36)
where p and q are parameters. This is best explained via a simple example. We choose f (x, y) =
A + By 3 , x2
where A and B are constants. Applying the right side of Eq. (4.36) yields λ−1 f (λ p x, λq y) =
A λ2 p+1 x 2
+ Bλ3q−1 y 3
p=−1/2,q=1/3
=
A + By 3 = f (x, y). x2
This also works for f (x, y) = Ax −2 y 3 . However it does not work for 7
See Widom (1965).
4.2 Beyond Van Der Waals Theory
141
f (x, y) =
A + By 3 , 1 + x2
because λ−1 f (λ p x, λq y) =
A + Bλ3q−1 y 3 λ + λ2 p+1 x 2
p=−1/2,q=1/3
=
A + By 3 . λ + x2
Therefore we do not expect that this idea applies to thermodynamic functions in general, except close to the critical point, where they may be expressed in terms of powers of δT and δ P. But what can we learn from Eq. (4.36)? We apply this equation to the free energy, i.e. (4.37) f ∞ (δt, δv) = λ−1 f ∞ (λ p δt, λq δv). The index ∞ is a reminder that we stay close to the critical point and we consider the leading part of the free energy in the sense discussed above. The exponent p should not be confused with the reduced pressure used in the universal van der Waals equation. With the particular choices (a) λ = δv −1/q and (b) λ = δt −1/ p we transform Eq. (4.37) into (a) :
f ∞ (δt, δv) = δv 1/q f ∞ (δv − p/q δt, 1)
(b) :
f ∞ (δt, δv) = δt
1/ p
f ∞ (1, δt
−q/ p
(4.38)
δv).
We want to use this to work out C V = −T ∂ 2 F/∂ T 2 |V and κT−1 = −V ∂ 2 F/ ∂ V 2 |T , i.e. using (a):C V = δv 1/q−2 p/q C˜ V (δv − p/q δt, 1) using
(b):κT−1
= δt
1/ p−2q/ p
κ˜ T (1, δt
−q/ p
(4.39)
δv).
Along the coexistence curve we had defined the exponent β via δv ∼ δt β . We use this to eliminate δv from the Eq. (4.39). Together with C V ∼ δt −α and κT ∼ δt −γ we obtain the following equations relating the exponents: −α =β
p 1 −2 , q q
−γ = −
1 q +2 , p p
β=
q . p
(4.40)
The third equation ensures that the scaling functions C˜ V and κ˜ T are “well behaved” at the critical point, i.e. δv − p/q δt approaches a constant at the critical point. We may eliminate p and q from these equations, which yields the critical exponent relation α + 2β + γ = 2.
(4.41)
142
4 Simple Phase Diagrams
Fig. 4.13 Temperature dependence of the isochoric heat capacity of sulfurhexafluoride near the critical point
The van der Waals values from Table 4.1 obviously satisfies this relation. Conversely we can use this relation to justify α = 0! Figure 4.13 shows C V measured for sulfurhexafluoride (S F6 ) along the critical isochore copied with permission from Haupt and Straub (1999) (S F6 : Tc = 318.7K , Pc = 37.6bar , Vc = 200cm 3 /mol).8 The value of the critical exponent α determined from these data is α = 0.1105+0.025 −0.027 . The currently accepted theoretical value is α = 0.110 ± 0.003 (Sengers and Shanks 2009). For β and γ the accepted theoretical values are β = 0.326±0.002 and γ = 1.239±0.002 in agreement with experiments and with the exponent relation Eq. (4.41). Note that these exponents do not depend on the molecular details of the fluid systems. Theoretical arguments show that the critical exponent values depend on space dimension, (order parameter) symmetry, and range of interaction (“short” versus “long”) but not on the details of molecular interaction. This allows to define so called universality classes of fluids (or near critical physical systems in general) with identical exponent values.9 It is easy to derive a critical exponent relation involving the exponent δ. Applying P = −∂ F/∂ V |T to (b) in Eq. (4.38) yields ˜ δt −q/ p δv). P = δt 1/ p−q/ p P(1,
(4.42)
Again with δv ∼ δt β along the coexistence curve we find βδ = 1/ p − q/ p or
8 On the critical isochore we have f (δt, δv = 0) = δt 1/ p f (1, 0) and therefore ∞ ∞ C V ∼ δt 1/ p−2 ∼ δt −α . That is the exponent is the same as on the coexistence curve. 9 It is a hypothesis that all fluids belong to one and the same universality class. A discussion of this hypothesis may be found in the above article by Sengers and Shanks.
4.2 Beyond Van Der Waals Theory
143
γ = β(δ − 1).
(4.43)
And again this relation is fulfilled by the van der Waals exponent values in Table 4.1. It is worth emphasizing that exponent relations like Eqs. (4.41) and (4.43) and others mainly serve to unify the picture, i.e. the number of independent critical exponents is greatly reduced. Already we have mentioned that the van der Waals exponents are incorrect. This incorrectness is not “just” due to the modest quantitative predictive power of the van der Waals approach. Here the latter misses the underlying physical picture completely. Every thermodynamic quantity fluctuates around an average value. Usually the fluctuations can be ignored entirely. This is what the van der Waals model does too. However, close to the critical point fluctuations become increasingly important and dominate over the average values.10 There are many models which in this respect are like the van der Waals model (Kadanoff 2009). When these models describe critical points, they all yield the same critical exponents—the so called mean field critical exponents.11 This is striking, because the models look quite different indeed. Nevertheless, near their respective critical points they all posses the same “symmetry”. Only after a method, the so called renormalization group, was invented how to properly deal with the dominating fluctuations in the critical region, was it possible to actually calculate the correct values for the critical exponents—and they still obey the above relations together with others derived via thermodynamic scaling.12 This is because of the great generality of the thermodynamic laws and, in addition, the fact that the power law behavior of thermodynamic functions near criticality turns out to be an integral part of the new theory as well.
4.2.1 The Clapeyron Equation Along the gas-liquid transition line in Fig. 4.14 we always have at “1” and “2” μl (1) = μg (1)
and
μl (2) = μg (2)
or
μl (2) − μl (1) = μg (2) − μg (1).
If “1” and “2” are infinitesimally close we may write dμl = −sl dT + vl d P = −sg dT + vg d P = dμg .
(4.44)
10 More precisely what happens is that the local fluctuations influence each other over large distances.
These distances are measured in terms of the fluctuation correlation length which diverges at the critical point. 11 In models for magnetic systems δ P is replaced by the corresponding magnetic field variable and δv is replaced by the magnetization. The compressibility is therefore replaced by the magnetic susceptibility. 12 A nice reference including historical developments is Fisher (1998).
144
4 Simple Phase Diagrams
Fig. 4.14 Thermodynamic paths on either side of the saturation line
P liquid 2 1
gas
T
Here lower case letters indicate molar quantities. Consequently we find sg − sl d P
= ,
dT coex vg − vl
(4.45)
where d P/dT is the slope of the gas-liquid transition line in Fig. 4.14. This equation is more general of course, because it applies not only to the transition from gas to liquid and vice versa but to the transition between any two phases we choose to call I and II. We remark that a transition with a non-zero latent heat, i.e. T s = 0, we call a first order phase transition.13 Thus we have d P
sI I − sI =
dT coex vI I − vI
(3.107),(3.165)
=
1 hI I − hI . T vI I − vI
(4.46)
This is the Clapeyron equation. Example: Enthalpy of Vaporization for Water. Here we calculate the enthalpy of vaporization, vap h, for water from the saturation pressure data shown in Fig. 4.4. Using Eq. (4.46) we may write d P
1 vap h ≈ .
dT coex T vgas
(4.47)
Compared to the molar volume of the gas we may neglect the liquid volume. If in addition vgas is expressed via the ideal gas law Eq. (4.47) becomes
13
Transitions without such discontinuity, e.g. at the gas-liquid critical point, are called continuous or (generally) second order.
4.2 Beyond Van Der Waals Theory
145
Fig. 4.15 Phase coexistence lines in the T-P-plane
P III
?
II
I
T
d ln P ≈
vap h dT R T2
(4.48)
or, after integration, ln
vap h P ≈− Po R
1 1 − T To
.
(4.49)
Using the values P = 4.246k Pa at T = 303.15K and P = 0.6113k Pa at T = 273.15K from the aforementioned figure, we obtain vap h = 44.5k J/mol in very good accord with vap h = 45.05k J/mol at T = 273.15K or vap h = 44.0k J/mol at T = 298.15K taken from HCP. One important application of the Clapeyron equation is the following. Whereas the van der Waals theory only describes the transition between gas and liquid, we know that already a one component system may exhibit other phases—like the solid state. A sketch of the situation is shown in Fig. 4.15. There are transition lines (solid lines) separating phases I and II as well as II and III. The two lines may come together at (?) to form what is called a triple point. 14 How would this look like, i.e. how would we draw the line separating phases I and III? Can the dashed lines be correct? According to Eq. (4.46) we have at the triple point h I →I I dP h I I →I I I + h I I I →I h I →I I I = Tt = = dT v I →I I v I I →I I I + v I I I →I v I →I I I
1− 1−
h I I →I I I h I →I I I v I I →I I I v I →I I I
The second equality follows via
14
According to the phase rule this is the most complicated case in a one-component system.
.
146
4 Simple Phase Diagrams
(a)
(b) P
P III
III
II
II
I
I
T
T
Fig. 4.16 Alternative phase diagrams near a triple point
h I →I I + h I I →I I I + h I I I →I = 0 v I →I I + v I I →I I I + v I I I →I = 0, corresponding to a path enclosing the triple point in infinitesimal proximity (Both functions are state functions!). Because, according to our assumption, the slopes of the coexistence lines I–II and I–III are identical, we must require (1−...)/(1−...) = 1 and thus h I I →I I I h I →I I I = . v I →I I I v I I →I I I This means, according to Eq. (4.46), that the slopes of the two solid lines in Fig. 4.15 should coincide close to the triple point. This is not a satisfactory result. We conclude that the slopes of all three lines must be different near the triple point.15 This leaves us with the two alternatives depicted in Fig. 4.16. In alternative (a) the broken lines correspond to the continuation of the coexistence lines between phases I and II and phases I and III. In particular the shaded area is a region in which phase I is unstable with respect to II. On the other hand in the same area phase II is less stable than III. According to the solid lines, however, phase I is the most stable, which clearly is inconsistent. Thus we discard alternative (a). Alternative (b) does not suffer from this problem and is the correct one. We conclude that the continuation of the coexistence line between any two of the phases must lie inside the third phase. With this information we may now sketch out the phase diagram of a simple one component system shown in Fig. 4.17. There are three projections of course. The T -P-projection is what we just have talked about. Here G means gas, F means liquid and K means solid. According to the van der Waals theory the gas-liquid coexistence line should terminate in a critical point (C). We do not posses any knowledge of 15
Note that this conclusion based on the Clapeyron equation does not hold in cases when there are transitions involved without discontinuities h or v.
4.2 Beyond Van Der Waals Theory
147
Fig. 4.17 Phase diagram of a simple one-component system
whether or not the liquid-solid line terminates similarly.16 All three lines meet in the triple point. The remaining projections contain areas of phase coexistence due to the volume discontinuity at the transitions. It is worth noting that even a one-component system exhibits a much more complicated phase diagram, i.e. here we always concentrate on partial phase diagrams. Figure 4.18 shows an extended but still partial phase diagram for water (data sources: lower graph—HCP; upper graph—Martin Chaplin (http://www.lsbu.ac.uk/water/ phase.html)). Even though things already get complicated, the rules we have established thus far for phase diagrams are always satisfied. The special temperatures are the freezing temperature at 1 bar, T f , the boiling temperature at 1 bar, Tb , the critical temperature, Tc (together with the critical pressure, Pc ), as well as the triple point 16
However, gas and liquid differ in no essential aspect of order or symmetry. This clearly sets them apart from the crystal. We can choose a path in the T -P-plane leading us from the gas to the liquid phase without crossing a phase boundary, i.e. very smoothly. Based on this concept we would not expect to find a liquid-solid critical point.
148
4 Simple Phase Diagrams P [bar] 8000 VI
7000 6000
V
5000 II
4000
liquid water
3000 III 2000 Ih 1000 200
300
400
500
600
700
1000 Pc=220.6 bar 100 ice 10
liquid water
Ih
1 water vapor 0.1 Tf = 273.15 K
Tc=647.14K
Tb =373.15K
Pt =
0.01
0.006 bar Tt =273.16 K 200
300
T [K] 400
Fig. 4.18 Partial phase diagram for water
Fig. 4.19 A phase diagram in the T-H-plane
500
600
700
4.2 Beyond Van Der Waals Theory
149
temperature, Tt (together with the triple point pressure, Pt ). Roman numerals in the upper graph distinguish different high pressure ice phases. Example: Superconductor Thermodynamics. Figure 4.19 shows a phase diagram in the T -H -plane, i.e. in this system the magnetic H -field assumes the role of P. The two phases are named s and n. We easily can work out a version of the Clapeyron equation in this case. Our starting point is Eq. (4.44), i.e. v v (4.50) − ss dT − Bs · d H = −sn dT − Bn · d H . 4π 4π
Here we have used the mapping from (P, −V ) to ( H , v B/(4π )) described on p. 25 in the context of the discussion of Eq. (1.51). Notice that now v is a constant molar volume of the material to which the phase diagram in Fig. 4.19 applies. Analogous to Eq. (4.45) we find in the present case d H
v ( Bn − B s ) · = ss − sn .
4π dT coex
(4.51)
Our phase diagram is meant to apply to a type I superconductor. The letters s and n label the superconducting and the normal conducting phases, respectively. In the s-region we have therefore B s = 0. If in addition we use the linear relation B n = μr H , where μr is the magnetic permeability, then Eq. (4.51) becomes v d H 2
μr = ss − sn . (4.52)
8π dT coex Based on this equation and some additional information we may work out the coexistence line. The additional information consists of the empirical approximations to the molar heat capacities in superconducting and normal conducting phases at low temperatures, i.e. cs = aT 3 and cn = bT 3 + γ T (an example may be found in Chap. 33 of Ashcroft and Mermin (1976)). The quantities a, b, and γ are constants, which may by obtained via suitable experimental data. By simply integrating the thermodynamic relation c = T ∂s/∂ T | H from zero temperature to T we obtain ss (T ) = ss (0) + (a/3)T 3 and sn (T ) = sn (0) + (b/3)T 3 + γ T (It may be puzzling that cn is used below Tc . Here the normal phase can be produced by a weak magnetic field destroying the superconducting state with little effect on the heat capacity.). If we invoke what is called the third law of thermodynamics, i.e. s(0) = 0, which we discuss in Chap. 5, we can integrate Eq. (4.52). The result, after some algebra, is
150
4 Simple Phase Diagrams
T2 Hc (T ) = Hc (0) 1 − 2 , Tc
where Hc (0) =
2π γ Tc and Tc = vμr
3γ . a−b
(4.53)
(4.54)
Notice that the latent heat, T s, vanishes in the two end-points of the transition line. Even though thermodynamics by itself does not explain superconductivity, it does allow additional predictions provided that certain input is available. However, if this input does consist of approximations then the additional predictions will be approximate as well.
Phase Separation in the RPM Arguably the most important insight provided by van der Waals theory is the role of intermolecular interaction on gas-liquid phase separation. The latter requires shortranged repulsion as well as attraction. Here we discuss an overall neutral system consisting of charges +q and −q possessing hard core excluded volume. This model system is termed the restricted primitive model (RPM). Corresponding experimental systems are molten salts (Weiss 2010; Pitzer 1990). Does such a system posses a critical point analogous to the gas-liquid critical point in the van der Waals theory? A priori there is no easy answer, because there is no obvious net attractive interaction as in the latter theory.17 The free enthalpy of our model system is G = n + μ+ + n − μ− .
(4.55)
The indices refer to the two charge types. We approximate the chemical potentials via 1 N Aq2 1 n± N A − . (4.56) μ± ≈ μo,± + RT ln V 2 λ D 1 + b/λ D The first term describes contributions to the chemical potentials not dependent on ion concentration. The concentration dependence here enters through an ideal gas
One may argue that the immediate neighborhood of +q on average contains an excess of −q and that this leads to the net attraction. But this is a truly complicated system and such arguments should always be backed up by calculation.
17
4.2 Beyond Van Der Waals Theory
151
term, the second term in Eq. (4.56), and via the interaction between the ions in the framework of the Debye-Hückel theory when the ions posses the radius b, cf. (3.127), the third term in Eq. (4.56). Does this yield a critical point? We insert Eq. (4.56) into Eq. (4.55) and the result into Eq. (2.183). The dotted line in the left panel in Fig. 4.4 is the spinodal line obtained in the van der Waals theory. On the spinodal line the compressibility diverges and thus ∂G/∂ V |T,n 1 ,n 2 ,... = 0, cf. (2.183). Straightforward differentiation of our present G, using λ D ∝ V 1/2 , leads to 1 x =1 4T ∗ (1 + x)2
(4.57)
employing this condition. Here T ∗ = b RT /(N A q 2 ) and x = b/λ D . If we insert x = 1 + δx into Eq. (4.57), we obtain to leading order in the small quantity δx 1 − δx 2 = 16T ∗ .
(4.58)
We recognize that there are always two solutions x = 1 ± δx with the same T ∗ . The two solutions coincide if x = xc = 1 corresponding to T ∗ = Tc∗ = 1/16. The conclusion is that the RPM possesses a gas-liquid spinodal curve, here worked out in the vicinity of the critical point at Tc =
1 N Aq2 16 Rb
and
ρc = 2cc =
1 . 64π b3
(4.59)
Note that presently q 2 = e2 , and ρ is the total ion number concentration. Notice also that the replacement q 2 → q 2 /(4π εo εr ) yields Tc in SI-units. The pressure follows via integration of ∂ P
1 ∂G
= ,
∂ V T,±n V ∂ V T,±n i.e.
N Aq2 PV =1+ (n + + n − )RT RT b
ln[1 + x] 1 2+x . − x2 2x 1 + x
(4.60)
(4.61)
The resulting critical compressibility factor is Pc = 16 ln 2 − 11 ≈ 0.09035. ρc RTc
(4.62)
The above references list experimental critical parameters. In particular we may compare the compressibility factor Eq. (4.62), because it is a pure number. It turns out that the above model does not make quantitative predictions—except in selected cases. But for us it provides a valuable exercise. In fact the model is not wrong—it is incomplete. It turns out that association of ions into aggregates is the most important ingredient for a more accurate description of phase separation in molten salts and
152
4 Simple Phase Diagrams
related ionic systems. A detailed account of this can be found in Levin and Fisher (1996).
Electric Field Induced Critical Point Shift The following discussion of the electric field induced shift of the critical temperature, density, and pressure is not so much motivated by its practical importance but rather by the rich content of conceptual and technical aspects making this a valuable exercise. In the preceding section we have used Eq. (2.183), i.e. the divergence of the compressibility, to locate the critical point. Here we consider an ordinary dielectric
where E is the (macroscopic) liquid (dielectric constant εr ) in an electric field E, average electrical field in the liquid. We also may want to apply Eq. (2.183) to deduce the electric field effect on the location of the critical point. We must know, however,
or at constant E.
whether to work out the partial derivative at constant constant D We can find the answer via the following inequality ∂2 f ∂2 f − ∂ D 2 ∂ρ 2
∂2 f ∂ D∂ρ
2 > 0.
(4.63)
Here f = f (T, ρ, D) is the free energy density, depending on (constant) tempera (appropriture, T , particle density, ρ, and the magnitude of the displacement field, D ate in an isotropic medium).18 Inequality Eq. (4.63) expresses the requirement that f is convex in terms of ρ and D. A sign change signals a “dent” in f causing phase separation (cf. the 1D situation depicted in Fig. 4.2). Thus, the replacement of > 0 by = 0 in Eq. (4.63) yields the critical point, i.e ∂μ
1 ∂ E
∂μ
−
4π ∂ D T,ρ ∂ρ T,D ∂ D T,ρ
∂ E
∂ρ T,D
(A.3) ∂ E
= − ∂ D ∂∂ρD
T,ρ
= 0.
(4.64)
T,E
Making use of Eq. (A.1) we obtain the desired result valid at the critical point, i.e.19 ∂μ
=0
∂ρ Tc ,E
(4.65)
18 On p. 59 we had found F = F(T, D).
The density (volume) dependence was ignored, because
and thus D
and E are
= r E, it did not play a significant role in the example. In addition we use D
and E.
along the same direction. Moreover we can use the magnitudes instead of D 19 On p. 25 we had discussed situations in which one can obtain new thermodynamic relations via replacement of P through, for instance, the electric field strength, E. In the present case we could have applied this to the chemical stability condition in Eq. (3.16) to immediately obtain Eq. (4.65).
4.2 Beyond Van Der Waals Theory
153
We apply this formula to the free energy density f = fo +
1 E · D. 4π
(4.66)
Here f o is the part of the free energy density which does not depend on the field explicitly. The attendant chemical potential is μ = μo +
1 ∂
E D . T,D 4π ∂ρ
(4.67)
Differentiation at constant E and using D = εr (ρ)E yields
∂μ
∂μo
1 ∂ ∂
E D
=
+ T,D T,E ∂ρ T,E ∂ρ T 4π ∂ρ ∂ρ
2 ∂μo
1 2 ∂ εr (ρ)
E =
−
. ∂ρ T 8π ∂ρ 2 T,E
(4.68)
We can express dμo in terms of dT and d P as usual
and thus
1 S dμo = − dT + d P ρ ρ
(4.69)
∂μo
1 ∂ P
.
=
∂ρ T ρ ∂ρ T,E=0
(4.70)
We put everything together by combining Eqs. (4.65), (4.68), and (4.70), i.e. 0=
1 ∂ P(ρc )
1 2 ∂ 2 εr (ρc )
E − .
Tc ,E ρc ∂ρ Tc ,E=0 8π ∂ρ 2
(4.71)
Note that we first take the derivatives with respect to ρ under the indicated constraints and subsequently evaluate the result at ρc . Notice also that the first term does not vanish even though we have learned that 0 = ∂ P/∂ V |Tc = −(ρ/V )∂ P/∂ρ|Tc in the context of van der Waals theory. This is because the critical point we study now is for a certain field strength E = 0. However, we may expand the first term as follows 1 ∂ P(ρc )
∂ P(Tc,o + δT, ρc,o + δρ)
1 =
E=0 ρc ∂ρ Tc ,E=0 ρc,o + δρ ∂ρ δρ ∂ P(Tc,o , ρc,o )
1 1− ≈
E=0 ρc,o ρc,o ∂ρ =0
154
4 Simple Phase Diagrams
+
≈
∂ 2 P(Tc,o , ρc,o )
∂ 2 P(Tc,o , ρc,o )
δρ + δT
E=0 E=0 ∂ρ 2 ∂ T ∂ρ 1 ρc,o
=0 2 ∂ P(Tc,o , ρc,o )
∂ T ∂ρ
E=0
δT.
(4.72)
The index c, o indicates that this quantity is taken at the critical point of the same system in absence of the electric field (Tc = Tc,o + δT ; ρc = ρc,o + δρ). The first two terms in the square brackets are zero, because they are evaluated for vanishing field strength at the attendant critical point. Analogously we have −
1 2 ∂ 2 εr (ρc )
1 2 ∂ 2 εr (Tc,o , ρc,o )
E E ≈ .
2 Tc ,E E=0 8π ∂ρ 8π ∂ρ 2
(4.73)
Notice that E also is a small quantity, so that the right side is the leading term of the expansion. Combination the last two equations yields δT ≈
∂ 2 P(T , ρ )
1 ∂ 2 εr (Tc,o , ρc,o )
c,o c,o
ρc,o E 2 .
E=0 E=0 8π ∂ρ 2 ∂ T ∂ρ
(4.74)
This is the leading contribution to the field induced temperature shift for small field strength. In order to obtain δρ to the same order we must work from ∂ 2 μ/∂ρ 2 |Tc ,E = 0. The result is
2 ∂ P(Tc,o ,ρc,o )
∂ 3 P(Tc,o ,ρc,o )
1 K (εr ) + ρc,o −
2 ∂ T ∂ρ ∂ T ∂ρ E=0 E=0
δρ ≈ δT, (4.75) 3 ∂ P(Tc,o ,ρc,o )
3 ∂ρ E=0
2 ∂ 3 εr (Tc,o ,ρc,o )
∂ εr (Tc,o ,ρc,o )
where K (εr ) = . The shift of the critical pres
3 2 ∂ρ ∂ρ E=0 E=0 sure is simply ∂ P(Tc,o , ρc,o )
δT. (4.76) δP ≈
E=0 ∂T Notice that the shifts all are quadratic in the field strength.20
20
We remark that the pressure derivatives can be estimated using the van der Waals equation of state (∂ 2 P/∂ T ∂ρ|c = 6Z c , ∂ 3 P/∂ T ∂ρ 2 |c = 6Z c /ρc ; ∂ 3 P/∂ρ 3 |c = 9Z c /ρc , where Z c = 3/8 is the critical compressibility factor. A sufficiently accurate estimate of the dielectric constant derivatives is more difficult. Considering a permanent point dipole, μ,
in a spherical cavity inside a continuous dielectric medium characterized by a dielectric constant, εr , Onsager (Onsager (1936); Nobel prize in chemistry for his work on irreversible thermodynamics, 1968) has derived the following simple approximation (εr − 1)(2εr + 1)/εr = 4π μ2 N A ρ/(RT ), which may in principle be used for this purpose. We leave this to the interested reader.
4.3 Low Molecular Weight Mixtures
155
4.3 Low Molecular Weight Mixtures 4.3.1 A Simple Phenomenological Model for Liquid-liquid Coexistence The van der Waals approach is applicable to gas-liquid phase separation in a onecomponent system. Another type of phase separation is observed in binary mixtures. Depending on thermodynamic conditions the components may be miscible or not. A simple model describing this is based on the following molar free enthalpy approximation (l) (l) (l) (l) (l) (l) (l) (l) (4.77) g = x A g A + x B g B + x A ln x A + x B ln x B + χ x A x B . Here g A = μ∗A (l)/RT and g B = μ∗B (l)/RT are the reduced molar free enthalpies of two pure liquid components A and B. Mixing A and B gives rise to the mixing (l) free enthalpy described by the ln-terms. Note that the mole fractions are x A and (l) (l) x B = 1 − x A . We had obtained this contribution earlier, cf. (3.30), for mixtures of ideal gases. Here we consider liquids. Nevertheless we still assume ideal behavior. The last term is new. It introduces an additional interaction free enthalpy proportional to the two mole fractions. The quantity χ is a parameter in this theory. Figure 4.20 shows g for different values of the χ -parameter. If χ is less than a (l) (l) critical value then g is a convex function of x A (or x B ). This situation is analogous to the free energy in the van der Waals theory for temperatures above the critical (l) temperature. If χ = χc then the curvature of g at x A = 1/2 becomes zero. For still larger values of χ a “bump” develops - again analogous to the free energy in the van der Waals theory for temperatures less than the critical temperature. Driven by the second law the system now lowers its free enthalpy by separating into two types of regions, which over time will coagulate into two large domains, one depleted of A and one en
